🧠:<think>Okay, so I'm trying to figure out how to model this problem where I need to allocate a budget to block pathways in a network to minimize the spread of an invasive species. Let's break it down step by step.First, the region is represented as a graph G(V, E), where each vertex is a zone and each edge is a pathway. Each edge has a weight w(e) which represents how easy it is for the species to move along that pathway. The goal is to block some of these edges with a given budget B to minimize the expected spread.In the first part, I need to develop a mathematical model. So, the cost to block an edge e is proportional to its weight, specifically w(e) * c, where c is a constant. The probability of the species spreading through a pathway is inversely proportional to the amount spent on blocking it. Hmm, so if I spend more on blocking an edge, the probability decreases.Let me think about how to model the probability. If the probability is inversely proportional to the amount spent, maybe it's something like P(e) = k / x(e), where x(e) is the amount spent on edge e, and k is some constant. But since we're dealing with a budget, maybe we can normalize it or incorporate it into the model without needing the constant.Wait, actually, since the cost is w(e)*c, and the amount spent on blocking edge e is x(e), which is a portion of the budget. So, maybe the probability of the species spreading through edge e is proportional to 1/(x(e)/w(e)). Because if you spend more relative to the weight, the probability decreases. Alternatively, it could be 1/(x(e)), but considering the weight, perhaps it's better to normalize by the weight.Alternatively, maybe the probability is inversely proportional to the amount spent, so P(e) = 1 / x(e). But since the cost is w(e)*c, the amount spent x(e) is proportional to w(e). So, maybe the probability is inversely proportional to x(e)/w(e), which would be 1/(x(e)/w(e)) = w(e)/x(e). That makes sense because if you spend more relative to the weight, the probability decreases.But I'm not entirely sure. Maybe I should think in terms of the expected number of pathways the species can take. The expected spread could be the sum over all edges of the probability that the species uses that edge. So, the total expected spread is the sum of P(e) for all edges e.Given that, the problem becomes minimizing the sum of P(e) over all edges e, subject to the total cost being less than or equal to B. The total cost is the sum over all edges of x(e), where x(e) is the amount spent on edge e, and x(e) = w(e)*c * t(e), where t(e) is the fraction of the edge's weight that we're blocking. Wait, no, actually, the cost to block edge e is x(e) = w(e)*c * something. Maybe x(e) is the amount spent, which is proportional to w(e). So, perhaps x(e) = c * w(e) * t(e), where t(e) is a binary variable indicating whether we block the edge or not. But if we can partially block edges, then t(e) could be a continuous variable between 0 and 1.Wait, the problem says "each edge can be blocked at a cost proportional to its weight w(e) × c". So, maybe the cost to block edge e is w(e)*c, and if we block it, we pay that cost. But if we don't block it, we pay nothing. But that would make it a binary decision: block or not block each edge. However, the probability of spreading is inversely proportional to the amount spent. If we can only block or not block, then the amount spent is either 0 or w(e)*c. So, if we block it, the probability is 0, and if we don't block it, the probability is 1/(w(e)*c). But that doesn't seem right because the probability should be inversely proportional to the amount spent, which is either 0 or w(e)*c.Wait, that might not make sense because if we don't block it, the amount spent is 0, and the probability would be infinite, which isn't practical. So perhaps instead, the amount spent can be a continuous variable, and the probability is inversely proportional to the amount spent. So, if we spend x(e) on edge e, then the probability of the species spreading through e is k / x(e), where k is a constant. But since we can choose x(e) between 0 and some maximum, perhaps the maximum being w(e)*c, but I'm not sure.Alternatively, maybe the probability is inversely proportional to the amount spent relative to the weight. So, P(e) = 1 / (x(e)/w(e)) = w(e)/x(e). That way, if you spend more relative to the weight, the probability decreases.But I'm getting confused. Let's try to formalize it.Let me denote x(e) as the amount spent on blocking edge e. The cost to block edge e is proportional to its weight, so x(e) = c * w(e) * t(e), where t(e) is the fraction of the edge's weight that we're blocking. But t(e) can be between 0 and 1, meaning we can partially block the edge. The probability of the species spreading through edge e is inversely proportional to the amount spent on it, so P(e) = k / x(e). But since we can choose x(e), we can set t(e) to control P(e).But since we want to minimize the expected spread, which is the sum of P(e) over all edges, we need to minimize sum_{e} P(e) = sum_{e} k / x(e). Subject to the total cost sum_{e} x(e) <= B.But we can also express x(e) in terms of t(e): x(e) = c * w(e) * t(e). So, substituting, the expected spread becomes sum_{e} k / (c * w(e) * t(e)).We need to minimize this sum subject to sum_{e} c * w(e) * t(e) <= B, and 0 <= t(e) <= 1 for all e.This is starting to look like an optimization problem. Let's write it formally.Minimize sum_{e in E} [k / (c * w(e) * t(e))] subject to sum_{e in E} [c * w(e) * t(e)] <= B and 0 <= t(e) <= 1 for all e.But since k and c are constants, we can simplify the objective function. Let's let k/c be another constant, say, m. Then the objective becomes sum_{e} [m / (w(e) * t(e))].But actually, since k and c are constants, we can ignore them for the purpose of optimization because they don't affect the t(e) values, only the scale of the objective. So, we can consider minimizing sum_{e} [1 / (w(e) * t(e))].But this seems a bit abstract. Maybe we can think of it differently. The expected spread is the sum over all edges of the probability of the species using that edge. If we block an edge, the probability decreases. So, the more we block, the less the probability.But the problem is that we have a budget constraint. So, we need to decide how much to spend on each edge to minimize the total expected spread.Wait, another approach: if the probability of spreading through edge e is inversely proportional to the amount spent on it, then P(e) = k / x(e). The total expected spread is sum_{e} P(e) = sum_{e} k / x(e). We need to minimize this sum subject to sum_{e} x(e) <= B and x(e) >= 0.But in this case, x(e) is the amount spent on edge e, which is proportional to w(e)*c. Wait, no, the cost to block edge e is x(e) = c * w(e) * t(e), where t(e) is the fraction blocked. So, the amount spent is x(e) = c * w(e) * t(e), and the probability is P(e) = k / x(e) = k / (c * w(e) * t(e)).So, the total expected spread is sum_{e} [k / (c * w(e) * t(e))]. We need to minimize this sum subject to sum_{e} [c * w(e) * t(e)] <= B, and 0 <= t(e) <= 1.But since k and c are constants, we can factor them out. Let's let m = k / c, then the objective becomes sum_{e} [m / (w(e) * t(e))]. So, we can ignore m for optimization purposes because it's a constant multiplier.Thus, the problem reduces to minimizing sum_{e} [1 / (w(e) * t(e))] subject to sum_{e} [w(e) * t(e)] <= B / c, and 0 <= t(e) <= 1.But since B is given, and c is a constant, we can let B' = B / c, so the constraint becomes sum_{e} [w(e) * t(e)] <= B'.So, the optimization problem is:Minimize sum_{e} [1 / (w(e) * t(e))]Subject to:sum_{e} [w(e) * t(e)] <= B'0 <= t(e) <= 1 for all e.This is a convex optimization problem because the objective is convex in t(e) and the constraint is linear.To solve this, we can use Lagrange multipliers. Let's set up the Lagrangian:L = sum_{e} [1 / (w(e) * t(e))] + λ (sum_{e} [w(e) * t(e)] - B')Take the derivative of L with respect to t(e) and set it to zero:dL/dt(e) = -1 / (w(e) * t(e)^2) + λ w(e) = 0Solving for t(e):-1 / (w(e) * t(e)^2) + λ w(e) = 0=> λ w(e) = 1 / (w(e) * t(e)^2)=> λ = 1 / (w(e)^2 * t(e)^2)But this must hold for all e, which implies that 1 / (w(e)^2 * t(e)^2) is the same for all e. Let's denote this common value as λ.So, for all e, t(e) = 1 / (w(e) * sqrt(λ))But we also have the constraint sum_{e} [w(e) * t(e)] <= B'Substituting t(e):sum_{e} [w(e) * (1 / (w(e) * sqrt(λ)))] = sum_{e} [1 / sqrt(λ)] = n / sqrt(λ) <= B'Where n is the number of edges. Wait, no, that can't be right because each term is 1 / sqrt(λ), so summing over all edges gives n / sqrt(λ). But n is the number of edges, which is fixed.Wait, but actually, t(e) must be <=1, so 1 / (w(e) * sqrt(λ)) <=1 => sqrt(λ) >= 1 / w(e). So, sqrt(λ) >= max(1 / w(e)) over all e.But this seems conflicting because if we set t(e) = 1 / (w(e) * sqrt(λ)), and we have sum_{e} [w(e) * t(e)] = sum_{e} [1 / sqrt(λ)] = (number of edges) / sqrt(λ) <= B'So, sqrt(λ) >= number of edges / B'But also, sqrt(λ) >= max(1 / w(e))So, sqrt(λ) is the maximum of these two values.But this seems complicated. Maybe there's a better way.Alternatively, since the problem is convex, the optimal t(e) will be such that the marginal cost of reducing t(e) is equal across all edges. That is, the derivative of the objective with respect to t(e) should be proportional to the derivative of the constraint.From the derivative, we have:-1 / (w(e) * t(e)^2) + λ w(e) = 0=> λ = 1 / (w(e)^2 * t(e)^2)So, for all e, λ is the same, meaning that 1 / (w(e)^2 * t(e)^2) is constant across e. Therefore, t(e) is proportional to 1 / w(e). Specifically, t(e) = k / w(e), where k is a constant.But since t(e) must be <=1, we have k <= w(e) for all e. So, k is the minimum of w(e). Wait, no, because t(e) = k / w(e) <=1 => k <= w(e). So, k must be <= min(w(e)).But we also have the budget constraint:sum_{e} [w(e) * t(e)] = sum_{e} [w(e) * (k / w(e))] = sum_{e} k = n * k <= B'So, k <= B' / nBut k must also be <= min(w(e))Therefore, k = min(B' / n, min(w(e)))Wait, but this might not be the case because the optimal t(e) could be such that some edges are fully blocked (t(e)=1) and others are not.Alternatively, perhaps the optimal solution is to set t(e) proportional to 1 / w(e), but scaled such that the total cost equals B'.Let me think again. If we set t(e) = k / w(e), then the total cost is sum_{e} [w(e) * t(e)] = sum_{e} [w(e) * (k / w(e))] = sum_{e} k = n * k = B'So, k = B' / nBut we also need t(e) <=1, so k / w(e) <=1 => w(e) >= k = B' / nSo, for edges where w(e) >= B' / n, t(e) = B' / (n * w(e)) <=1For edges where w(e) < B' / n, t(e) would be greater than 1, which is not allowed. So, in that case, we set t(e)=1 for those edges, and allocate the remaining budget to the other edges.Wait, this is getting somewhere. So, the strategy is:1. Identify edges with w(e) <= B' / n. For these edges, set t(e)=1 (fully block them), and subtract their cost from B'.2. For the remaining edges, set t(e) proportional to 1 / w(e), scaled so that the total cost equals the remaining budget.But this might be a bit involved. Alternatively, perhaps the optimal t(e) is such that for edges with higher w(e), we spend less, and for edges with lower w(e), we spend more, but not exceeding the budget.Wait, actually, since the probability is inversely proportional to the amount spent, we want to spend more on edges where the probability would be higher if we don't block them. But higher w(e) means easier spread, so we might want to block those more.But the cost to block is proportional to w(e), so blocking a high w(e) edge is more expensive. So, there's a trade-off: high w(e) edges are more important to block because they have higher spread potential, but they are also more expensive to block.So, perhaps the optimal strategy is to block the edges with the highest w(e) first, up to the budget.But wait, in the first part, without considering vulnerability, the model would be to block edges in a way that the marginal reduction in expected spread per unit cost is equal across all edges.From the Lagrangian, we saw that the optimal t(e) is proportional to 1 / w(e). So, the fraction of the budget spent on each edge is inversely proportional to its weight.But let's think about it differently. The expected spread is sum_{e} P(e) = sum_{e} k / x(e). The cost is sum_{e} x(e) = B.To minimize the expected spread, we need to allocate the budget such that the marginal reduction in expected spread per unit cost is the same across all edges.The marginal reduction in expected spread for edge e is dP(e)/dx(e) = -k / x(e)^2. The marginal cost is 1 per unit x(e). So, the ratio of marginal reduction to marginal cost is -k / x(e)^2. To equalize this across all edges, we set -k / x(e)^2 = λ for all e, which implies x(e) = sqrt(k / λ). But since k and λ are constants, x(e) is the same for all edges. Wait, that can't be right because x(e) should depend on w(e).Wait, no, because the cost to block edge e is x(e) = c * w(e) * t(e). So, the amount spent is proportional to w(e). So, if we set x(e) proportional to 1 / w(e), then t(e) would be proportional to 1 / w(e)^2.Wait, let's try again. The marginal reduction in expected spread per unit cost for edge e is:dP(e)/dx(e) / (d cost / dx(e)) = (-k / x(e)^2) / 1 = -k / x(e)^2We want this to be equal across all edges, so:-k / x(e)^2 = λ for all e=> x(e) = sqrt(k / λ)But this suggests that x(e) is the same for all edges, which contradicts the idea that higher w(e) edges are more expensive to block.Wait, perhaps I'm missing something. The cost to block edge e is x(e) = c * w(e) * t(e). So, the derivative of cost with respect to t(e) is c * w(e). Therefore, the marginal cost per unit t(e) is c * w(e).The marginal reduction in expected spread per unit t(e) is dP(e)/dt(e) = dP(e)/dx(e) * dx(e)/dt(e) = (-k / x(e)^2) * (c * w(e)).So, the ratio of marginal reduction to marginal cost is:[(-k / x(e)^2) * (c * w(e))] / (c * w(e)) = -k / x(e)^2Which again suggests that this ratio is the same across all edges, leading to x(e) being equal for all edges. But this can't be right because x(e) = c * w(e) * t(e), so if x(e) is equal, then t(e) is inversely proportional to w(e).So, t(e) = x(e) / (c * w(e)) = constant / w(e)Therefore, the fraction of the budget spent on each edge is inversely proportional to its weight.So, the optimal strategy is to allocate the budget such that the amount spent on each edge is proportional to 1 / w(e). But since the cost to block edge e is proportional to w(e), this means that the fraction of the budget spent on edge e is proportional to 1 / w(e).Wait, let me clarify. If we set x(e) proportional to 1 / w(e), then since x(e) = c * w(e) * t(e), we have t(e) = x(e) / (c * w(e)) proportional to 1 / w(e)^2.But this might not be the most straightforward way to think about it.Alternatively, perhaps the optimal allocation is to spend the budget such that the product of the amount spent and the weight is constant across edges. That is, x(e) * w(e) = constant.But I'm getting tangled up here. Let's try to think of it in terms of the Lagrangian again.We have:dL/dt(e) = -1 / (w(e) * t(e)^2) + λ w(e) = 0=> λ = 1 / (w(e)^2 * t(e)^2)So, for all e, 1 / (w(e)^2 * t(e)^2) is equal. Let's call this common value λ.Therefore, for all e, w(e)^2 * t(e)^2 = 1 / λ=> t(e) = sqrt(1 / (λ w(e)^2)) = 1 / (sqrt(λ) w(e))So, t(e) is inversely proportional to w(e).Now, the total cost is sum_{e} [w(e) * t(e)] = sum_{e} [w(e) / (sqrt(λ) w(e))] = sum_{e} [1 / sqrt(λ)] = n / sqrt(λ) = B'So, sqrt(λ) = n / B'Therefore, t(e) = 1 / ( (n / B') w(e) ) = B' / (n w(e))But t(e) must be <=1, so B' / (n w(e)) <=1 => w(e) >= B' / nSo, for edges where w(e) >= B' / n, t(e) = B' / (n w(e)) <=1For edges where w(e) < B' / n, t(e) would be greater than 1, which is not allowed. So, for those edges, we set t(e)=1, and adjust the allocation for the remaining edges.Wait, this is a key point. So, the optimal allocation is:- For edges with w(e) < B' / n, set t(e)=1 (fully block them)- For edges with w(e) >= B' / n, set t(e) = B' / (n w(e))But we need to ensure that the total cost doesn't exceed B'.Let me calculate the total cost:Total cost = sum_{e: w(e) < B'/n} [w(e) * 1] + sum_{e: w(e) >= B'/n} [w(e) * (B' / (n w(e)))] = sum_{e: w(e) < B'/n} w(e) + sum_{e: w(e) >= B'/n} (B' / n)Let’s denote S = sum_{e: w(e) < B'/n} w(e)And let m be the number of edges with w(e) >= B'/nThen, total cost = S + m * (B' / n)But we need this to be <= B'So, S + (m B') / n <= B'=> S <= B' (1 - m / n)But S is the sum of weights of edges with w(e) < B'/nThis seems a bit abstract. Maybe it's better to proceed step by step.1. Sort all edges in increasing order of w(e).2. Start with the smallest w(e). Since w(e) is small, B'/n might be larger than w(e), so we set t(e)=1 for these edges until we reach an edge where w(e) >= B'/n.3. For edges with w(e) >= B'/n, we set t(e) = B' / (n w(e))But we need to ensure that the total cost doesn't exceed B'Alternatively, perhaps the optimal solution is to block all edges with w(e) <= k, where k is chosen such that the total cost of blocking those edges plus the cost of partially blocking the remaining edges equals B'.This is similar to a knapsack problem where we decide which edges to fully block and which to partially block.But this is getting quite involved. Maybe I should look for a more straightforward approach.Another way to think about it is that the expected spread is minimized when the marginal reduction in spread per unit cost is equal across all edges. So, for each edge, the derivative of the expected spread with respect to the cost spent on it should be equal.From earlier, we have:dP(e)/dx(e) = -k / x(e)^2And the marginal cost is 1 per unit x(e). So, the ratio is -k / x(e)^2.To equalize this across all edges, we set x(e) equal for all edges. But since x(e) = c * w(e) * t(e), this implies that t(e) is inversely proportional to w(e).So, t(e) = constant / w(e)But t(e) must be <=1, so the constant is the minimum of w(e) * t(e) such that t(e)=1 for some edges.Wait, perhaps the optimal solution is to set t(e) = min(1, B' / (n w(e)))But I'm not sure. Maybe I should look for an example.Suppose we have two edges, e1 with w(e1)=1 and e2 with w(e2)=2. Suppose B'=3.If we set t(e1)=1, cost=1, remaining budget=2.Then, t(e2)=2 / (2 * 2)=0.5Total cost=1 + 2=3Expected spread= k / (1 *1) + k / (2 *0.5)= k + k=2kAlternatively, if we set t(e1)=3/(2*1)=1.5>1, so set t(e1)=1, cost=1, remaining budget=2.Then, t(e2)=2/(2*2)=0.5Same as before.Alternatively, if B'=4, then t(e1)=4/(2*1)=2>1, so set t(e1)=1, cost=1, remaining budget=3.t(e2)=3/(2*2)=0.75Total cost=1 + 2*0.75=1 +1.5=2.5 <4. Wait, no, the total cost is 1 + 2*0.75=2.5, but B'=4, so we have remaining budget=1.5.Wait, this approach might not be correct.Alternatively, perhaps the optimal allocation is to set t(e)=1 for all edges until the budget is exhausted, but that might not be optimal because some edges are more critical.Wait, maybe I should think in terms of the reciprocal. Since the expected spread is sum_{e} 1/(x(e)), and x(e)=c w(e) t(e), we can write the expected spread as sum_{e} 1/(c w(e) t(e)).To minimize this, we need to maximize sum_{e} 1/(t(e)) subject to sum_{e} w(e) t(e) <= B'/c.But this is equivalent to minimizing sum_{e} 1/(t(e)) with sum_{e} w(e) t(e) <= B'This is a convex optimization problem, and the solution is t(e) proportional to 1/w(e). So, t(e)=k/w(e), where k is chosen such that sum_{e} w(e) * (k/w(e))=k n <= B'So, k=B'/nThus, t(e)=B'/(n w(e))But t(e) must be <=1, so for edges where w(e)>=B'/n, t(e)=B'/(n w(e))<=1For edges where w(e)<B'/n, t(e)=1So, the optimal strategy is:- For all edges e, if w(e) < B'/n, set t(e)=1 (fully block)- For edges e with w(e) >= B'/n, set t(e)=B'/(n w(e))This ensures that the total cost is sum_{e: w(e)<B'/n} w(e) + sum_{e: w(e)>=B'/n} (B'/n) <= B'Because sum_{e: w(e)>=B'/n} (B'/n) = m * (B'/n), where m is the number of such edges.And sum_{e: w(e)<B'/n} w(e) + m * (B'/n) <= B'This is the optimal allocation.Now, moving on to part 2, where each zone v has a vulnerability level α(v). We need to modify the model to prioritize zones based on their vulnerability.So, the expected impact is not just the number of edges the species can spread through, but also the vulnerability of the zones it reaches. So, if a zone is more vulnerable, the impact of the species spreading there is higher.Therefore, the expected impact should be a weighted sum, where the weight is the vulnerability of the destination zone.Wait, but the graph is undirected, so each edge connects two zones. So, the spread through edge e affects both zones connected by e. Therefore, the impact of blocking edge e should consider the vulnerability of both zones it connects.Alternatively, perhaps the expected impact is the sum over all edges of the probability of the species spreading through e multiplied by the vulnerability of the zones connected by e.But this is getting more complex. Let's think carefully.The expected impact can be modeled as the sum over all edges of the probability of the species spreading through e multiplied by the vulnerability of the zones it connects. But since each edge connects two zones, the impact could be the sum of the vulnerabilities of both zones.Alternatively, perhaps the impact is the sum over all zones of the vulnerability of the zone multiplied by the probability that the species reaches that zone.But modeling the probability of reaching each zone is more complicated because it depends on the entire network structure, not just individual edges.This seems more involved. Maybe a simpler approach is to consider that the impact of an edge is proportional to the product of the vulnerabilities of the two zones it connects. So, for edge e connecting u and v, the impact is α(u) * α(v) * P(e).Therefore, the total expected impact is sum_{e} α(u) α(v) P(e), where u and v are the endpoints of e.Given that, we need to minimize sum_{e} α(u) α(v) P(e) subject to sum_{e} x(e) <= B, where x(e) = c w(e) t(e), and P(e) = k / x(e).But similar to part 1, we can set up the optimization problem.Let me denote for edge e connecting u and v, the weight is w(e), and the impact coefficient is α(u) α(v).Then, the expected impact is sum_{e} [α(u) α(v) * k / x(e)].We need to minimize this sum subject to sum_{e} x(e) <= B, and 0 <= t(e) <=1.Again, we can express x(e) = c w(e) t(e), so the expected impact becomes sum_{e} [α(u) α(v) * k / (c w(e) t(e))].Ignoring constants, the objective is sum_{e} [α(u) α(v) / (w(e) t(e))].To minimize this, we can use a similar approach as before. The marginal reduction in expected impact per unit cost should be equal across all edges.The derivative of the objective with respect to t(e) is -α(u) α(v) / (w(e) t(e)^2). The marginal cost is c w(e). So, the ratio is (-α(u) α(v) / (w(e) t(e)^2)) / (c w(e)) = -α(u) α(v) / (c w(e)^2 t(e)^2).We want this ratio to be equal across all edges, so:-α(u) α(v) / (c w(e)^2 t(e)^2) = λ for all e=> α(u) α(v) / (w(e)^2 t(e)^2) = constant=> t(e) = sqrt(α(u) α(v)) / (w(e) sqrt(λ))So, t(e) is proportional to sqrt(α(u) α(v)) / w(e)Therefore, the optimal t(e) is proportional to sqrt(α(u) α(v)) / w(e)This means that for edges connecting more vulnerable zones, we should spend more to block them, even if their weight is higher.So, the allocation strategy changes to prioritize edges that connect zones with higher vulnerability, adjusting for the weight of the edge.In summary, the optimal allocation in part 2 is to set t(e) proportional to sqrt(α(u) α(v)) / w(e), ensuring that the total cost does not exceed the budget B.This means that edges connecting highly vulnerable zones will have a higher fraction of their weight blocked, even if they have higher w(e), because the impact of their spread is greater.So, the key difference from part 1 is that in part 2, the allocation considers not just the weight of the edge but also the vulnerability of the connected zones, giving higher priority to blocking edges that could lead to higher impact invasions.</think>

🧠:<think>Okay, so I have this problem about designing a park with a circular pond and a rectangular garden. The pond has a radius of 30 meters. The garden is rectangular, with its length twice its width, and it shares a 40-meter side with the pond, tangentially along the circumference. I need to find the area of the rectangular garden and then the total area occupied by both the pond and the garden, considering the overlapping part where they touch tangentially.Let me start by visualizing the setup. There's a circular pond with radius 30 meters. The rectangular garden is next to it, sharing a 40-meter side. Since the garden is tangential to the pond, that means the side of the garden just touches the pond at exactly one point. So, the side of the garden is a tangent to the circle.First, I need to figure out the dimensions of the rectangular garden. It says the length is twice its width. Let me denote the width as 'w' and the length as '2w'. The problem mentions that the garden shares a 40-meter side with the pond. So, is this 40-meter side the length or the width of the garden?Wait, the garden is a rectangle, so it has two sides of each dimension. The side that is 40 meters is the one that is tangent to the pond. Since the garden is placed next to the pond, the 40-meter side must be the side that is tangent. But is this side the length or the width?Hmm, the problem says the garden is tangential along the pond's circumference. So, the side that is 40 meters is the one that is tangent. So, the 40-meter side is either the length or the width of the garden.But since the length is twice the width, if the 40-meter side is the length, then the width would be 20 meters. If it's the width, then the length would be 80 meters. I need to figure out which one it is.Wait, the garden is placed such that it shares a 40-meter side with the pond. So, the side that is 40 meters is adjacent to the pond. Since the pond is circular, the side of the garden that is tangent to the pond must be a straight line touching the pond at one point.Therefore, the 40-meter side is the side that is tangent to the pond. So, whether it's the length or the width, it's the side that is 40 meters. So, if the length is twice the width, then we have two possibilities:1. If the 40-meter side is the length, then the width is 20 meters.2. If the 40-meter side is the width, then the length is 80 meters.But which one is it? The problem says the garden is tangential along the pond's circumference. So, the side that is tangent is the 40-meter side. So, whether it's length or width, it's 40 meters. So, depending on that, the other dimension is either 20 or 80.Wait, maybe I can figure it out by considering the position of the garden relative to the pond.If the garden is placed such that its 40-meter side is tangent to the pond, then the distance from the center of the pond to the garden's side is equal to the radius of the pond, which is 30 meters.But wait, the tangent line is at a distance equal to the radius from the center. So, the distance from the center of the pond to the side of the garden is 30 meters.But the garden is a rectangle, so if the side is 40 meters, how does that relate to the distance from the center?Wait, perhaps I need to consider the position of the garden relative to the pond. Let me sketch a diagram in my mind.Imagine the circular pond with center O. The garden is a rectangle with one side AB of length 40 meters tangent to the pond at point T. The rectangle extends from AB to another side CD, which is parallel to AB. The length of the garden is the distance between AB and CD, which is the width if AB is the length, or the length if AB is the width.Wait, this is getting confusing. Let me clarify.Let me denote:- Let AB be the side of the garden that is tangent to the pond, with length 40 meters.- The garden is a rectangle, so the opposite side CD is also 40 meters.- The other two sides, BC and AD, are the other dimension, which is either the width or the length.Given that the length is twice the width, so if AB is the length, then BC would be the width, and AB = 2 * BC. If AB is the width, then BC would be the length, and BC = 2 * AB.But since AB is 40 meters, if AB is the length, then BC = 20 meters. If AB is the width, then BC = 80 meters.So, which is it? Is AB the length or the width?The problem says the garden shares a 40-meter side with the pond. It doesn't specify whether it's the length or the width. Hmm.Wait, maybe the side that is tangent is the longer side or the shorter side? It depends on how the garden is placed. If the garden is placed such that its longer side is tangent, then the length is 40 meters, and the width is 20 meters. Alternatively, if the shorter side is tangent, then the width is 40 meters, and the length is 80 meters.But how can we determine which is the case?Wait, perhaps the position of the garden relative to the pond can help. Since the garden is placed next to the pond, the side that is tangent is the side that is adjacent to the pond. So, the garden is either placed such that its longer side is next to the pond or its shorter side.But without more information, it's ambiguous. Hmm.Wait, maybe I can use the fact that the side is tangent to the pond. The distance from the center of the pond to the tangent side is equal to the radius, which is 30 meters.So, if I can model this, maybe I can find the other dimension.Let me consider the rectangle with side AB = 40 meters, tangent to the circle at point T. The center of the circle is O, and OT is perpendicular to AB, with OT = 30 meters.Now, the rectangle extends from AB to CD, which is parallel to AB. The distance between AB and CD is the other dimension of the rectangle, say, h.So, if AB is the length, then h is the width, and h = 20 meters. If AB is the width, then h is the length, and h = 80 meters.But how do we know which one it is?Wait, perhaps the rectangle is placed such that the side AB is tangent, and the other sides extend away from the pond. So, the distance from the center to the opposite side CD would be OT + h, but that might not necessarily be the case.Wait, no, because the sides BC and AD are perpendicular to AB and CD. So, the distance from the center to AB is 30 meters, and the distance from the center to CD would be 30 + h or 30 - h, depending on the direction.But since the garden is outside the pond, the distance from the center to CD would be 30 + h.But without knowing h, it's hard to say.Wait, maybe I can use the Pythagorean theorem here.If I consider the rectangle, with AB = 40 meters, and the distance from the center to AB is 30 meters. Then, the diagonal from the center to the corner of the rectangle can be calculated.Wait, let me think.Let me place the center of the pond at point O. The side AB is tangent to the circle at point T. So, OT is perpendicular to AB, with OT = 30 meters.Now, the rectangle has sides AB = 40 meters, and the other sides BC and AD of length h.So, the rectangle extends from AB to CD, which is parallel to AB, at a distance h from AB.Now, the distance from the center O to CD would be OT + h, but since CD is on the opposite side of AB, it's actually OT + h if the rectangle is extending away from the center.But wait, actually, the distance from O to CD is not necessarily OT + h because CD is not necessarily aligned along the same line as OT.Wait, maybe I need to model this with coordinates.Let me set up a coordinate system where the center of the pond is at (0, 0). The side AB is a horizontal tangent to the circle. The tangent at point T can be at (x, y). Since it's a horizontal tangent, the point T must be at (a, 30) or (a, -30). Let's assume it's at (a, 30) for simplicity.But wait, the tangent line is horizontal, so the radius at the point of tangency is vertical. So, the point T is either (30, 0) or (-30, 0). Wait, no, if the tangent is horizontal, then the radius must be vertical, so the point T is at (0, 30) or (0, -30). Hmm, actually, no, wait.Wait, if the tangent is horizontal, then the radius at the point of tangency is vertical. So, the point T has the same x-coordinate as the center, which is (0, 30) or (0, -30). So, the tangent line is horizontal, so it's either y = 30 or y = -30.But in the problem, the garden is placed such that it shares a 40-meter side with the pond. So, the side AB is 40 meters, which is the tangent line.So, if the tangent line is y = 30, then AB is a horizontal line at y = 30, with length 40 meters. So, AB extends from (-20, 30) to (20, 30), making it 40 meters long.Then, the rectangle extends downward from AB. The other sides are vertical, from (-20, 30) to (-20, 30 - h) and from (20, 30) to (20, 30 - h). So, the lower side CD is from (-20, 30 - h) to (20, 30 - h).But the problem is that the garden is a rectangle with length twice its width. So, if AB is the length, then AB = 40 meters, and the width is h, which is half of 40, so 20 meters. Alternatively, if AB is the width, then the length is 80 meters, which would be the vertical sides.Wait, but in this coordinate system, the length of AB is 40 meters, which is the horizontal side. The vertical sides are of length h. So, if the length is twice the width, then if AB is the length, h is the width, so h = 20 meters. If AB is the width, then the length is 80 meters, which would mean the vertical sides are 80 meters, but in our case, the vertical sides are h, so h = 80 meters.But in the coordinate system, the vertical sides are from y = 30 to y = 30 - h. So, if h is 20 meters, the lower side is at y = 10. If h is 80 meters, the lower side is at y = -50.But the problem is that the garden is placed next to the pond, so it's outside the pond. The pond has a radius of 30 meters, so the lower side of the garden at y = 10 is still within the pond's area if h = 20 meters, because the pond extends down to y = -30.Wait, no, the pond is a circle centered at (0,0) with radius 30, so it goes from y = -30 to y = 30. If the garden is placed above the pond, with AB at y = 30, then the garden extends downward into the pond if h is 20 meters, which would cause overlap.But the problem says the garden is tangential to the pond, so it should only touch at one point, meaning the garden is entirely outside the pond except for the tangent point.Therefore, if h = 20 meters, the garden would extend into the pond, which is not allowed. Therefore, h must be such that the garden is entirely outside the pond.Wait, so if h is 80 meters, then the lower side is at y = 30 - 80 = -50, which is outside the pond, since the pond only goes down to y = -30. So, in that case, the garden is entirely outside the pond except for the tangent point at (0, 30). That makes sense.Therefore, the garden has AB = 40 meters as the width, and the length is 80 meters, with the vertical sides being 80 meters each.Wait, but in the problem statement, it says the garden is a rectangle with length twice its width. So, if AB is the width, then the length is 80 meters, which is twice 40 meters. So, that fits.Therefore, the garden has width 40 meters and length 80 meters.Wait, but hold on. If AB is the width, then the length is the vertical side, which is 80 meters. But in our coordinate system, the vertical sides are of length h, so h = 80 meters. Therefore, the garden extends from y = 30 to y = -50, which is 80 meters.But the pond is only up to y = -30, so the garden extends 20 meters beyond the pond on the lower side.Therefore, the area of the garden is length times width, which is 80 meters * 40 meters = 3200 square meters.Wait, but let me double-check. If AB is the width, then the length is 80 meters, so the area is 40 * 80 = 3200 m².Alternatively, if AB is the length, then the width is 20 meters, but that would cause the garden to overlap with the pond, which is not acceptable. Therefore, AB must be the width, making the length 80 meters.So, the area of the garden is 3200 m².Now, moving on to the second part: determining the total area occupied by both the pond and the garden, considering the overlapping section where the garden is tangential to the pond.Wait, but if the garden is only tangent at one point, then the overlapping area is just a single point, which has zero area. Therefore, the total area would be the area of the pond plus the area of the garden.But that seems too straightforward. Let me think again.Wait, maybe the garden is placed such that it's adjacent to the pond along a side, but not just tangent at a single point. Wait, the problem says it's tangential along the pond's circumference, which implies that the side is tangent, meaning it touches at exactly one point. So, the overlapping area is just a point, which doesn't contribute to the area.Therefore, the total area occupied by both the pond and the garden is simply the sum of their areas.So, the area of the pond is πr² = π*(30)² = 900π square meters.The area of the garden is 3200 square meters.Therefore, the total area is 900π + 3200 square meters.But wait, let me confirm if the garden is entirely outside the pond except for the tangent point. If the garden is placed such that its lower side is at y = -50, which is outside the pond, then yes, there is no overlapping area except at the tangent point, which doesn't contribute to the area.Therefore, the total area is just the sum of the two areas.But let me think again. Maybe the garden is placed such that part of it is over the pond? But no, because it's tangential, meaning it just touches at one point, so the rest is outside.Therefore, the total area is 900π + 3200.But let me calculate the numerical value for better understanding.900π is approximately 900 * 3.1416 ≈ 2827.44 m².Adding 3200 m² gives approximately 2827.44 + 3200 ≈ 6027.44 m².But since the problem asks for the exact value, I should keep it in terms of π.So, the total area is 3200 + 900π square meters.Wait, but let me make sure I didn't make a mistake in determining the dimensions of the garden.Earlier, I considered AB as the width, making the length 80 meters. But let me verify this.If AB is the width, then the length is 80 meters. The garden extends from y = 30 to y = -50, which is 80 meters. The width is 40 meters, so the garden is 40 meters wide and 80 meters long.But wait, in the coordinate system, the garden is placed such that its width is along the x-axis from (-20, 30) to (20, 30), which is 40 meters. Then, it extends downward 80 meters to (-20, -50) and (20, -50).But the pond is a circle with radius 30, so it goes from y = -30 to y = 30. Therefore, the garden extends 20 meters below the pond, but the rest is outside.Therefore, the garden does not overlap with the pond except at the tangent point.Therefore, the total area is indeed the sum of the two areas.So, the area of the garden is 3200 m², and the total area is 3200 + 900π m².Wait, but let me think again about the dimensions.If AB is the width, then the length is 80 meters. But in the coordinate system, the length is the vertical side, which is 80 meters. So, the garden is 40 meters wide (along the x-axis) and 80 meters long (along the y-axis). So, the area is 40 * 80 = 3200 m².Yes, that seems correct.Alternatively, if AB were the length, then the width would be 20 meters, but that would cause the garden to extend into the pond, which is not allowed because it's only supposed to be tangent at one point.Therefore, AB must be the width, making the length 80 meters.So, the area of the garden is 3200 m², and the total area is 3200 + 900π m².Wait, but let me make sure about the tangent point.If the garden is placed such that its side AB is tangent to the pond at point T, which is at (0, 30), then the garden extends from (-20, 30) to (20, 30) along AB, and then down to (-20, -50) and (20, -50). So, the garden is a rectangle that is 40 meters wide and 80 meters long, placed such that its top side is tangent to the pond at (0, 30).Therefore, the garden does not overlap with the pond except at the tangent point, so the total area is the sum of the two areas.Therefore, the answers are:1. Area of the garden: 3200 m².2. Total area: 3200 + 900π m².But let me write them in the required format.1. The area of the rectangular garden is 3200 square meters.2. The total area occupied by both the pond and the garden is 3200 + 900π square meters.Alternatively, if we need to write it as a single expression, it's 900π + 3200 m².But I think the problem expects both areas to be considered, so the total area is the sum.Yes, that makes sense.So, to summarize:1. The garden's area is 40 meters * 80 meters = 3200 m².2. The total area is the pond's area (900π m²) plus the garden's area (3200 m²), totaling 3200 + 900π m².</think>

🧠:<think>Alright, so I've got this problem about a political analyst studying a geopolitical network using a directed graph. The nodes are countries, and the edges represent influence with weights. There are two parts to the problem, both related to analyzing this network. Let me try to unpack each part step by step.Starting with part 1: They want to determine the overall influence of each country using the PageRank algorithm. I remember that PageRank is a way to rank nodes in a graph based on the structure of the graph, particularly the links (or in this case, influence edges). The equation given is r = αP^T r + (1-α)v. So, I need to figure out how to set up the matrix equation and describe an iterative method to solve it.First, the PageRank formula. The vector r is the PageRank vector, α is the damping factor, P is the transition probability matrix, and v is the personalization vector. The equation is r = αP^T r + (1-α)v. So, to solve for r, we can rearrange this equation.Let me write it out:r = αP^T r + (1 - α)vTo solve for r, we can bring the αP^T r term to the left side:r - αP^T r = (1 - α)vFactor out r:r(I - αP^T) = (1 - α)vThen, solving for r:r = (I - αP^T)^{-1} (1 - α)vBut inverting a matrix is computationally expensive, especially for large n. So, instead, we use an iterative method. The standard approach is the power method, which iteratively multiplies the transition matrix until convergence.So, the iterative method would start with an initial guess for r, say r0, and then compute r1 = αP^T r0 + (1 - α)v, then r2 = αP^T r1 + (1 - α)v, and so on until the vector converges.But wait, I should think about how P is constructed. P is the transition probability matrix. Each entry P_ij represents the probability of moving from node j to node i. Since the graph is directed, each node j has out-degree equal to the number of countries it influences. The transition probability from j to i is w(j,i) divided by the sum of all weights from j.So, for each column j of P, the entries are P_ij = w(j,i) / sum_{k} w(j,k). That makes sense because it's a probability distribution over the outgoing edges from j.Therefore, to compute P, we need to normalize each column by the sum of the weights in that column. That way, each column sums to 1, making P a stochastic matrix.Now, the damping factor α is typically set around 0.85 in the original PageRank algorithm. It accounts for the probability that a random surfer continues following links versus jumping to a random page. Here, it's similar, but instead of jumping to a random page, it's using the personalization vector v. If v is uniform, it's like jumping to a random country; otherwise, it's jumping according to the distribution in v.So, the iterative method is straightforward:1. Initialize r0 as a vector of ones or some uniform distribution.2. For each iteration k, compute r^{(k)} = αP^T r^{(k-1)} + (1 - α)v.3. Check for convergence: if ||r^{(k)} - r^{(k-1)}|| < ε, stop; else, continue.This is the power method, and it's efficient because each iteration only requires a matrix-vector multiplication, which is O(n^2) time. However, for large n, even O(n^2) can be expensive, so optimizations like sparse matrix representations are necessary if the graph is sparse.Moving on to part 2: The analyst wants to find the top k pairs of countries with the highest mutual influence scores. The mutual influence score M(i,j) is defined as w(i,j) * w(j,i). So, for each pair (i,j), we compute the product of the weights in both directions.First, I need to think about how to compute this efficiently. The straightforward approach is to iterate over all possible pairs (i,j), compute M(i,j), and then select the top k. But the number of pairs is O(n^2), which can be very large if n is big. For example, if n is 100,000, n^2 is 10^10, which is way too big to handle directly.So, we need a way to compute the top k pairs without checking all possible pairs. Let me think about the structure of the graph. Each edge has a weight, and mutual influence requires both (i,j) and (j,i) edges to exist. So, first, we can filter out all pairs where either w(i,j) or w(j,i) is zero because their mutual influence would be zero.Therefore, the first step is to identify all pairs (i,j) where both edges exist. Then, for each such pair, compute M(i,j) and keep track of the top k.But even this can be expensive if the graph is dense. For a dense graph, the number of mutual pairs is still O(n^2). So, we need a way to find the top k without enumerating all mutual pairs.One approach is to use a heap data structure. We can iterate through each node i, and for each i, look at its outgoing edges. For each neighbor j of i, if there's a reciprocal edge from j to i, compute M(i,j) and add it to a max-heap. Then, extract the top k elements from the heap.But even this approach can be time-consuming if the graph is large because for each node, we have to check all its neighbors and then check for reciprocal edges. However, if the graph is represented as an adjacency list, and for each node, we have a set of its neighbors, we can quickly check if j is in i's neighbors and vice versa.Alternatively, we can precompute for each node i, a list of nodes j where both (i,j) and (j,i) exist, along with their mutual influence scores. Then, we can collect all these scores and find the top k.But again, for large n, this can be computationally intensive. So, perhaps we can use a more efficient method.Another idea is to note that the mutual influence score is the product of two weights. If we can find pairs where both weights are large, their product will be large. So, perhaps we can focus on nodes with high out-degrees or high-weight edges.But I'm not sure if that's necessarily the case because a pair with moderate weights could still have a high product. So, maybe it's better to think in terms of the top edges.Wait, perhaps we can use a selection algorithm. Since we only need the top k pairs, we don't need to compute all mutual influence scores. Instead, we can find the top k pairs by only considering the top edges.But how?Alternatively, we can use a priority queue approach where we process the edges in a certain order and keep track of the top k products.But I'm not sure. Maybe another approach is to realize that mutual influence is symmetric, so M(i,j) = M(j,i). Therefore, we can consider each unordered pair only once, which reduces the number of pairs by half. But in terms of computation, it's still O(n^2) in the worst case.Wait, but if the graph is sparse, then the number of mutual edges is much smaller. For example, if each node has an average degree d, then the number of mutual edges is roughly O(n*d^2), which could be manageable if d is small.But if the graph is dense, this approach isn't helpful.Alternatively, perhaps we can use matrix multiplication techniques. If we have the adjacency matrix A, where A_ij = w(i,j), then the matrix of mutual influence scores is A .* A^T, where .* is the element-wise product. Then, the diagonal elements of this matrix would be the mutual influence scores. But computing this matrix explicitly is O(n^3), which is not feasible for large n.Wait, no, actually, the mutual influence score matrix would be A multiplied element-wise with A transpose. So, M = A .* A^T. Then, the top k elements of M can be found by looking at the top k elements of this matrix.But computing M explicitly is O(n^2), which is again not feasible for very large n.So, perhaps we need a way to find the top k elements without constructing the entire matrix.One efficient way is to use a heap-based approach where we generate candidate pairs and keep track of the top k.Here's an idea: for each node i, we can sort its outgoing edges in descending order of weight. Similarly, sort the incoming edges for each node j. Then, for each node i, we can look at the top edges in its outgoing list and see if the corresponding nodes have high incoming edges from i.But this is still a bit vague.Alternatively, we can use a method similar to the one used in finding the top k elements in a matrix product. Since M is the element-wise product of A and A^T, we can think of it as a matrix where each element is the product of A_ij and A_ji.To find the top k elements, we can use a max-heap. We can iterate through each element A_ij, and if A_ji exists, compute the product and add it to the heap. However, this would require iterating through all elements, which is O(n^2), which is not feasible for large n.Wait, but if we can find a way to only consider the top candidates, perhaps we can limit the number of computations.Another approach is to use a priority queue where we keep track of the top candidates. For example, we can start by considering the top edges from each node and their reciprocals, adding their products to the heap, and then extracting the maximum each time until we have k elements.But I'm not sure about the exact implementation.Alternatively, perhaps we can use a two-pointer technique. For each node i, we have its outgoing edges sorted in descending order. Similarly, for each node j, incoming edges are sorted. Then, for each i, we can pair its top outgoing edge with the top incoming edge from that neighbor, compute the product, and keep track of the maximum.But this might not capture all possible top k pairs, as some pairs might not be the top for either node but still have a high product.Hmm, this is tricky. Maybe the best approach is to accept that for a general graph, finding the top k mutual influence scores requires O(n^2) time, but we can optimize it by only considering pairs where both edges exist and using efficient data structures to track the top k.So, the steps would be:1. For each node i, create a list of nodes j where both (i,j) and (j,i) edges exist. For each such pair, compute M(i,j) = w(i,j)*w(j,i).2. Collect all these M(i,j) scores.3. Sort them in descending order and pick the top k.But for large n, this is O(n^2), which is not feasible. So, we need a way to find the top k without enumerating all possible pairs.Wait, perhaps we can use a selection algorithm. If we can find the k largest elements in a set without sorting the entire set, that would be more efficient. But even so, we still need to generate the set of mutual influence scores, which is O(n^2).Alternatively, if we can represent the mutual influence scores in a way that allows us to find the top k without explicitly computing all of them, that would be better.Wait, another idea: since M(i,j) = w(i,j)*w(j,i), we can think of it as the product of two numbers. So, for each node i, we can look at its top outgoing edges and see if those nodes have high incoming edges from i.So, for each node i, we can take the top m outgoing edges (where m is a small number, say 100), and for each such j, check if w(j,i) is also large. If so, compute M(i,j) and add it to a candidate list. Then, after processing all nodes, we can collect all these candidates and find the top k.This way, we limit the number of pairs we consider to n*m, which is much smaller than n^2 if m is small. However, this approach might miss some pairs where both w(i,j) and w(j,i) are moderately large but not in the top m for either node.So, it's a trade-off between accuracy and computational efficiency. For the sake of finding the top k pairs, perhaps this approach is acceptable, especially if k is much smaller than n.Alternatively, we can use a more sophisticated method like the one used in the "top-k" element-wise product of two matrices. There's an algorithm called the "top-k" element-wise product which can find the top k elements without computing all of them.But I'm not sure about the exact method. Maybe it's similar to the approach used in the "top-k" singular values or something else.Wait, perhaps we can use a heap to keep track of the top candidates. Here's how it might work:1. For each node i, sort its outgoing edges in descending order of weight.2. For each node j, sort its incoming edges in descending order of weight.3. Initialize a max-heap. For each node i, take the top edge (i,j) and if w(j,i) exists, compute M(i,j) and add it to the heap.4. Extract the maximum M(i,j) from the heap, add it to the result list.5. For the extracted pair (i,j), take the next edge from i's outgoing list and compute M with j's incoming edge, add to the heap if it exists.6. Repeat until we've extracted k pairs.This way, we don't have to consider all pairs, only the top candidates, which can significantly reduce the number of computations.But implementing this requires careful management of the heap and ensuring that we don't miss any potential top pairs. It's similar to the approach used in the "k largest elements in a matrix" problem.So, in summary, the method would involve:- Preprocessing the graph to sort the outgoing and incoming edges for each node.- Using a max-heap to track the top mutual influence scores.- Iteratively extracting the maximum and adding new candidates until we've found the top k.This approach reduces the computational complexity from O(n^2) to something more manageable, especially if k is much smaller than n.But I should also consider the space complexity. Storing all the edges and their weights is necessary, so for a graph with m edges, it's O(m) space. The heap would store O(n) elements at most, which is manageable.Another optimization is to note that mutual influence is symmetric, so once we've considered pair (i,j), we don't need to consider (j,i) again. So, we can process each unordered pair only once.But in the heap approach, since we're considering each node's outgoing edges, we might end up processing both (i,j) and (j,i). To avoid duplication, we can enforce an order, say i < j, and only process pairs where i < j. This way, each pair is considered only once.So, incorporating that into the algorithm:1. For each node i, sort its outgoing edges in descending order of weight, considering only j > i to avoid duplicates.2. For each node j, sort its incoming edges in descending order of weight.3. Initialize the heap with the top edge from each i where j > i and w(j,i) exists.4. Extract the maximum M(i,j), add to the result.5. For the extracted pair (i,j), take the next edge from i's outgoing list (still j > i) and compute M(i,j) if w(j,i) exists, add to the heap.6. Continue until k pairs are extracted.This way, we ensure that each pair is processed only once, reducing the number of computations by half.In terms of computational complexity, if each node has an average degree d, the preprocessing step is O(m log m) for sorting. The heap operations are O(m log m) as well, since each edge is pushed and popped once. But since we're only interested in the top k, the actual number of operations might be less, depending on k.So, overall, the method is feasible for large networks as long as k is not too large, and the graph isn't excessively dense.To summarize, the steps for part 2 are:1. For each node i, sort its outgoing edges (j > i) in descending order of w(i,j).2. For each node j, sort its incoming edges in descending order of w(j,i).3. Use a max-heap to keep track of candidate pairs (i,j) where both edges exist, starting with the top edges from each i.4. Extract the maximum M(i,j) from the heap, add to the result.5. For each extracted pair, add the next candidate from i's outgoing list to the heap.6. Repeat until k pairs are found.This approach efficiently finds the top k mutual influence pairs without having to compute all possible pairs, making it suitable for large networks.Now, putting it all together, I think I have a good grasp of both parts. For part 1, it's about setting up the PageRank equation and using the power method iteratively. For part 2, it's about efficiently finding the top k mutual influence pairs using a heap-based approach after preprocessing the edges.I should make sure I didn't miss any key points. For part 1, the matrix equation is correctly formulated, and the iterative method is the power method. For part 2, the mutual influence score is correctly defined, and the optimization method using heaps and avoiding duplicate pairs is a solid approach.I think I'm ready to write the final answer now.</think>

🧠:<think>Okay, so I have this problem about modeling the number of notable film songs produced each year in Pakistan from 1950 to 2020. The function f(t) is piecewise, with an exponential part from 1950 to 1985 (which is t=0 to t=35) and a quadratic part from 1985 to 2020 (t=35 to t=70). I need to find the constants a, b, c, d, and e using the given data points: 20 songs in 1950, 100 in 1985, and 50 in 2020. Then, I have to calculate the total number of songs by integrating f(t) from 0 to 70.First, let's note the given information:- In 1950 (t=0), f(t) = 20.- In 1985 (t=35), f(t) = 100.- In 2020 (t=70), f(t) = 50.So, for the first part of the function, when t is between 0 and 35, f(t) = ae^{bt} + c. At t=0, this should equal 20. So plugging in t=0:f(0) = a*e^{0} + c = a*1 + c = a + c = 20.So, equation 1: a + c = 20.Next, at t=35, the function switches to the quadratic part. But since the function is continuous at t=35, the value from the exponential part should equal the value from the quadratic part at t=35. So, f(35) from the exponential side is ae^{35b} + c, and from the quadratic side, it's d*(35 - 35)^2 + e = d*0 + e = e. So, ae^{35b} + c = e.But we also know that f(35) is 100, so:Equation 2: ae^{35b} + c = 100.And equation 3: e = 100.Wait, hold on. Because in the quadratic function, at t=35, f(t) = e, which is 100. So e is 100.So, equation 3: e = 100.Then, moving on to the quadratic part. The quadratic function is f(t) = d(t - 35)^2 + e, and at t=70, f(t) = 50. So plugging in t=70:f(70) = d*(70 - 35)^2 + e = d*(35)^2 + e = 1225d + e = 50.But we know e is 100, so:Equation 4: 1225d + 100 = 50.So, solving for d:1225d = 50 - 100 = -50d = -50 / 1225 = -10/245 = -2/49 ≈ -0.0408163265.So, d = -2/49.Alright, so now we have e = 100, d = -2/49.Now, going back to the exponential part. We have two equations:1. a + c = 20.2. ae^{35b} + c = 100.We need another equation to solve for a, b, c. But wait, we only have two equations so far. Maybe we need another condition? Hmm.Wait, perhaps the function is differentiable at t=35? The problem doesn't specify, but sometimes in such models, they assume continuity and differentiability. Let me check.If the function is differentiable at t=35, then the derivatives from both sides should be equal.So, the derivative of the exponential part is f’(t) = abe^{bt}, and the derivative of the quadratic part is f’(t) = 2d(t - 35).At t=35, the derivative from the exponential side is abe^{35b}, and from the quadratic side is 2d*(0) = 0.So, if they are equal, then abe^{35b} = 0.But abe^{35b} = 0 implies either a=0, b=0, or e^{35b}=0. But e^{35b} is never zero, so either a=0 or b=0.If a=0, then from equation 1, c=20, and equation 2 becomes 0 + 20 = 100, which is not true. So a can't be zero.If b=0, then f(t) = a + c, which is a constant function. But from t=0 to t=35, f(t) would be constant, but at t=35, it jumps to 100, which is a discontinuity unless a + c = 100, but from t=0, a + c =20. So that can't be.Therefore, the function is not differentiable at t=35, or at least the problem doesn't specify it. So, we can't use that condition.So, we only have two equations for a, b, c:1. a + c = 20.2. ae^{35b} + c = 100.We need another equation. Maybe we can assume something else? Or perhaps the function is smooth in some other way? Hmm.Wait, maybe the problem expects us to have only two equations, but with three variables, so perhaps we need to make an assumption or find a relationship between a, b, c.Alternatively, maybe we can express c in terms of a from equation 1: c = 20 - a.Then, substitute into equation 2:a e^{35b} + (20 - a) = 100.Simplify:a e^{35b} + 20 - a = 100.a(e^{35b} - 1) = 80.So, a = 80 / (e^{35b} - 1).But we still have two variables here: a and b. So, unless we have another condition, we can't solve for both.Wait, maybe we can assume that the exponential growth rate is such that the function increases smoothly to 100 at t=35. But without another data point, I don't think we can determine both a and b.Wait, perhaps the problem expects us to model the exponential part with only two points, which is possible, but we need another condition. Maybe the function is increasing, so b is positive? Or maybe we can assume that the growth rate is such that it's a reasonable number.Alternatively, maybe I misread the problem. Let me check again.The function is piecewise: exponential from t=0 to t=35, quadratic from t=35 to t=70. Given f(0)=20, f(35)=100, f(70)=50.We have three points, but the function is piecewise, so each piece is defined over its interval.For the exponential part, we have two points: t=0 and t=35. So, we can solve for a, b, c.Wait, actually, for the exponential part, we have two equations:1. At t=0: a + c =20.2. At t=35: a e^{35b} + c =100.So, two equations with three variables: a, b, c. So, we need another condition. Maybe the function is differentiable? But earlier, that led to a problem.Alternatively, perhaps the quadratic part is also defined with another condition? Wait, the quadratic part is defined only at t=35 and t=70, but we have two points for the quadratic: t=35 (f=100) and t=70 (f=50). So, with two points, we can solve for d and e, which we did: d=-2/49, e=100.But for the exponential part, we have two points, but three variables. So, unless we have another condition, we can't solve for a, b, c uniquely.Wait, perhaps the problem is expecting us to model the exponential part with just two points, meaning we can express a and c in terms of b, but without another condition, we can't find numerical values.Hmm, maybe I made a mistake earlier. Let me think again.Wait, the function is piecewise, so from t=0 to t=35, it's exponential, and from t=35 to t=70, it's quadratic. So, the exponential part is defined on [0,35), and the quadratic on [35,70]. So, at t=35, the function is defined by the quadratic part, which is 100, so the exponential part must also approach 100 as t approaches 35 from the left.So, we have:At t=0: a + c =20.At t=35: a e^{35b} + c =100.So, two equations, three unknowns. So, unless we have another condition, like the derivative at t=35, which we tried earlier but it led to a contradiction unless a=0 or b=0, which isn't feasible.Alternatively, maybe the problem expects us to set c=0? But that would make f(0)=a=20, and then f(35)=20 e^{35b}=100, so e^{35b}=5, so b=(ln5)/35≈0.043. But the problem didn't specify c=0, so that might not be valid.Alternatively, maybe we can assume that the exponential function passes through another point? But we only have two points: t=0 and t=35.Wait, perhaps the problem is designed such that the exponential part is only defined up to t=35, and the quadratic part takes over. So, maybe we can model the exponential part with just two points, which would require assuming c=0? But that's not given.Alternatively, maybe the problem expects us to use the fact that the quadratic part is a parabola opening downward because d is negative, which we found. So, the vertex is at t=35, which is the maximum point, which makes sense because the number of songs peaks at 100 in 1985 and then decreases to 50 in 2020.But for the exponential part, it's increasing from 20 to 100 over 35 years. So, it's a growth curve.But without another condition, I can't find a unique solution for a, b, c. Maybe the problem expects us to set c=0, but that's an assumption.Wait, let me try that. If c=0, then from equation 1: a=20.Then, equation 2: 20 e^{35b}=100 => e^{35b}=5 => 35b=ln5 => b=(ln5)/35≈0.043.So, a=20, b≈0.043, c=0.But is c=0 a valid assumption? The problem didn't specify, but in the function f(t)=ae^{bt}+c, c is the baseline. If c=0, then the function starts at a=20, which is given. So, maybe that's acceptable.Alternatively, if c≠0, then we have a different scenario. For example, if c=10, then a=10, and then 10 e^{35b} +10=100 => 10 e^{35b}=90 => e^{35b}=9 => 35b=ln9≈2.1972 => b≈0.0628.So, different a and b.But without another condition, we can't determine a and b uniquely.Wait, maybe the problem expects us to model the exponential part as a simple growth without an additive constant, meaning c=0. That would make sense because otherwise, we have too many variables.So, perhaps the intended solution is to set c=0, then a=20, and b=(ln5)/35.Alternatively, maybe the problem expects us to have c=20, but that would make a=0, which doesn't make sense because then f(t)=20 for all t, which contradicts f(35)=100.So, perhaps c=0 is the way to go.Alternatively, maybe the problem expects us to have a continuous derivative, but as we saw earlier, that leads to abe^{35b}=0, which is not possible unless a=0 or b=0, which isn't feasible.So, perhaps the problem expects us to ignore the differentiability and just use the two points to solve for a and b, assuming c=0.Alternatively, maybe the problem is designed such that c=20, but then a=0, which doesn't work.Wait, perhaps I can express a and c in terms of b.From equation 1: c=20 - a.From equation 2: a e^{35b} + (20 - a) =100.So, a(e^{35b} -1)=80.Thus, a=80/(e^{35b} -1).So, a is expressed in terms of b.But without another equation, we can't find a unique solution.Wait, maybe the problem expects us to use the fact that the quadratic part is a parabola, and perhaps the exponential part is a simple growth curve, so maybe we can choose b such that the growth is reasonable.Alternatively, maybe the problem expects us to have c=20, but then a=0, which doesn't work because f(35)=0 +20=20≠100.So, perhaps the only way is to set c=0, a=20, and then find b.So, let's proceed with that assumption.So, a=20, c=0.Then, from equation 2: 20 e^{35b}=100 => e^{35b}=5 => 35b=ln5 => b=(ln5)/35≈0.043.So, b≈0.043.So, now, we have:For the exponential part: f(t)=20 e^{0.043t}.For the quadratic part: f(t)= (-2/49)(t -35)^2 +100.Now, let's verify if this makes sense.At t=0: f(0)=20 e^0=20. Correct.At t=35: f(35)=20 e^{35*0.043}=20 e^{1.505}≈20*4.505≈90.1. Wait, but we need f(35)=100. Hmm, that's a problem.Wait, because if a=20, c=0, then f(35)=20 e^{35b}=100 => e^{35b}=5 => 35b=ln5≈1.6094 => b≈0.046.Wait, I think I miscalculated earlier.Wait, 35b=ln5≈1.6094, so b≈1.6094/35≈0.046.So, b≈0.046.So, f(t)=20 e^{0.046t}.At t=35: 20 e^{1.6094}=20*5=100. Correct.So, that works.So, a=20, b≈0.046, c=0.Wait, but earlier, when I set c=0, I thought f(35)=20 e^{35b}=100, which gives b=(ln5)/35≈0.046.So, that works.So, perhaps the problem expects us to set c=0, which gives a=20, and then solve for b.So, let's go with that.So, constants are:a=20,b=(ln5)/35≈0.046,c=0,d=-2/49≈-0.0408,e=100.So, that's the solution.Now, moving on to part 2: calculating the total number of notable film songs produced from 1950 to 2020 by evaluating the integral of f(t) from 0 to70.So, the integral is the sum of two integrals: from 0 to35 of ae^{bt} dt, plus from35 to70 of d(t-35)^2 + e dt.So, let's compute each integral.First integral: ∫₀³⁵ 20 e^{0.046t} dt.The integral of e^{kt} dt is (1/k)e^{kt} + C.So, ∫20 e^{0.046t} dt = (20 /0.046) e^{0.046t} + C.So, evaluated from 0 to35:(20 /0.046)(e^{0.046*35} - e^0) = (20 /0.046)(e^{1.61} -1).We know that e^{1.61}≈5.00 (since ln5≈1.6094), so e^{1.61}≈5.00.So, approximately:(20 /0.046)(5 -1)= (20 /0.046)(4)= (20*4)/0.046≈80 /0.046≈1739.13.But let's compute it more accurately.First, 0.046 is approximately ln5 /35≈0.0459815.So, 20 /0.0459815≈435.07.Then, e^{0.0459815*35}=e^{1.60935}=5.So, exactly, it's (20 / (ln5 /35))*(5 -1)= (20 *35 / ln5)*4= (700 / ln5)*4.Compute 700 / ln5: ln5≈1.6094, so 700 /1.6094≈435.07.Then, 435.07*4≈1740.28.So, the first integral is approximately1740.28.Second integral: ∫₃₅⁷⁰ [(-2/49)(t -35)^2 +100] dt.Let's make a substitution: let u = t -35. Then, when t=35, u=0; when t=70, u=35.So, the integral becomes ∫₀³⁵ [(-2/49)u² +100] du.Compute this:∫ [(-2/49)u² +100] du = (-2/49)*(u³/3) +100u +C.Evaluate from 0 to35:[ (-2/49)*(35³/3) +100*35 ] - [0 +0] = (-2/49)*(42875/3) +3500.Compute 35³=42875.So, (-2/49)*(42875/3)= (-2/49)*(42875)/3.Compute 42875 /49: 49*875=42875, so 42875 /49=875.So, (-2/49)*(42875/3)= (-2)*875 /3= (-1750)/3≈-583.333.Then, adding 3500: 3500 -583.333≈2916.667.So, the second integral is approximately2916.667.Therefore, total integral≈1740.28 +2916.667≈4656.947.So, approximately4657 notable film songs produced from1950 to2020.But let's compute it more accurately.First integral:(20 / (ln5 /35))*(5 -1)= (20*35 / ln5)*4= (700 / ln5)*4.Compute 700 / ln5: ln5≈1.6094379124341003.So, 700 /1.6094379124341003≈435.0703.Then, 435.0703*4≈1740.2812.Second integral:(-2/49)*(35³/3) +100*35.35³=42875.So, (-2/49)*(42875/3)= (-2/49)*(42875)/3.42875 /49=875.So, (-2)*875 /3= (-1750)/3≈-583.3333333.100*35=3500.So, total second integral:3500 -583.3333333≈2916.6666667.Total integral:1740.2812 +2916.6666667≈4656.9478667.So, approximately4656.95.So, rounding to the nearest whole number, approximately4657 notable film songs.But let's check the exact value.First integral:(20 / (ln5 /35))*(5 -1)= (20*35 / ln5)*4= (700 / ln5)*4.700 / ln5=700 /1.6094379124341003≈435.0703.435.0703*4≈1740.2812.Second integral:(-2/49)*(35³/3) +100*35= (-2/49)*(42875/3) +3500= (-2/49)*(42875)/3 +3500.42875 /49=875.So, (-2)*875 /3= (-1750)/3≈-583.3333333.3500 -583.3333333≈2916.6666667.Total≈1740.2812 +2916.6666667≈4656.9478667.So, approximately4656.95.So, the total number of notable film songs produced from1950 to2020 is approximately4657.But let's see if we can express it exactly.First integral:(20 / (ln5 /35))*(5 -1)= (20*35 / ln5)*4= (700 / ln5)*4.Second integral:(-2/49)*(35³/3) +100*35= (-2/49)*(42875/3) +3500= (-2*42875)/(49*3) +3500= (-85750)/147 +3500.Simplify:-85750 /147= -583.3333333.So, total second integral:3500 -583.3333333=2916.6666667.So, total integral:(700 / ln5)*4 +2916.6666667.But 700 / ln5 is exact, so we can write it as:Total= (2800 / ln5) +2916.6666667.But 2916.6666667 is 8750/3.So, Total=2800 / ln5 +8750/3.We can leave it like that, but probably the problem expects a numerical value.So, 2800 / ln5≈2800 /1.60943791243≈1740.2812.8750 /3≈2916.6667.So, total≈1740.2812 +2916.6667≈4656.9479.So, approximately4657.Therefore, the total number of notable film songs produced from1950 to2020 is approximately4657.But let me double-check the integrals.First integral: ∫₀³⁵20 e^{0.046t} dt.We can write it as (20 /0.046)(e^{0.046*35} -1).0.046*35=1.61.e^{1.61}=5.00 (approximately, since ln5≈1.6094).So, e^{1.61}=5.00.So, (20 /0.046)(5 -1)= (20 /0.046)*4≈(434.7826)*4≈1739.13.Wait, earlier I had 1740.28, but this is slightly different.Wait, because 0.046 is an approximation of ln5 /35≈0.0459815.So, 20 / (ln5 /35)=20*35 / ln5≈700 /1.6094≈435.07.Then, 435.07*(5 -1)=435.07*4≈1740.28.So, that's more accurate.So, the first integral is approximately1740.28.Second integral: ∫₃₅⁷⁰ [(-2/49)(t -35)^2 +100] dt.As we computed, it's2916.6667.So, total≈1740.28 +2916.67≈4656.95.So, approximately4657.Therefore, the total number of notable film songs produced from1950 to2020 is approximately4657.So, to summarize:Constants:a=20,b=ln5 /35≈0.0459815,c=0,d=-2/49≈-0.0408163,e=100.Total notable film songs≈4657.</think>