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给一棵树，每个节点是一座城市，每次有两种操作，1炸掉一座城市花费x，2炸掉一座城市及其所在的联通块内...

作成日: 2025年9月14日

使用モデル GPT-5 by Chat01

質問

给一棵树，每个节点是一座城市，每次有两种操作，1炸掉一座城市花费x，2炸掉一座城市及其所在的联通块内的所有城市，花费y，求最大花费

思考

Maximizing destruction costs

The user is asking about a strategy for destroying cities in a tree-like structure to maximize cost. You can choose operations like destroying a single city at cost x or destroying a whole connected component at cost y. To maximize the total cost, you could mix strategies: destroy cities individually first, then use a component-destruction operation, if it benefits the overall cost.

Maximizing destruction through partitioning

In a tree, destroying a city (operation 1) splits its component into multiple parts. Then, operation 2 can be applied to each part. The goal is to maximize the total cost by choosing some cities for operation 1, which increases the number of components based on the nodes' degrees. After removing cities, you can calculate the resulting number of components.

Maximizing total cost strategy

When removing cities individually, the cost is m*x. Then, for each resulting component, we use operation 2 to remove it entirely at cost y per component. The total cost is m*x + C*y, where C is the number of components. To maximize, the strategy depends on the sum of the degrees of the removed nodes, and whether x - y is positive or negative.

Adjusting removal strategy

The formula for the total cost involves the sum of degree-based weights for each removed node, with the costs scaled by y and x - y. But, removing adjacent nodes changes their degrees, so the approach might need adjustment. In a tree, the final number of components is determined by the number of connected components in the remaining forest, factoring in the earlier removals.

Calculating components after node removal

When I remove nodes, the number of components in the tree depends on how many edges get removed and how the remaining nodes are connected. Initially, the formula seemed fine, but when testing cases like a path of four nodes, I realized I needed to account for isolated components formed after deletions. After adjustments, the cost and component formula seems to align better.

Optimizing node selection

To maximize the cost, I need to choose a set S that maximizes the combined effects of node removals and component creation. This involves balancing edge cuts, component penalties, and node contributions. The formula suggests maximizing the boundary edges (|δ(S)|), minimizing internal components (c_S), and correctly accounting for each node's contribution. This can be approached using dynamic programming on trees, where decisions at each node depend on its connection to its child or its parent.

Optimizing dynamic programming solution

I’m working through the logic for choosing the best set S in a tree to maximize total cost. The algorithm considers nodes in different states based on connections to their parent. The recurrence relations are carefully structured to compute the best subset of nodes while accounting for edge cuts, component creation, and node contributions. I’ll need to adjust some interpretations for edge cases and ensure the final computation yields the correct result.

Thought for 1m 49s

回答

下面这题的关键在于：
先用若干次“单点爆破”（操作1，费用 x）删掉一部分点，使树被切成若干联通块；最后对每个剩余联通块各用一次“整块爆破”（操作2，费用 y）。要最大化总费用。

把被单点删掉的点集记作 SS。删完 SS 后剩下的联通块数 CC 满足（树上恒等式）：

C  =  1+∣δ(S)∣+eS−∣S∣C \;=\; 1 + |\delta(S)| + e_S - |S|

其中 ∣δ(S)∣|\delta(S)| 是跨越割 SS 与补集的边数，eSe_S 是 SS 内部的边数。
于是总费用：

cost(S)=∣S∣⋅x+C⋅y=y  +  (∣S∣⋅x+y⋅∣δ(S)∣−y⋅cS)⏟F(S)\text{cost}(S)= |S|\cdot x + C\cdot y = y \;+\; \underbrace{\big(|S|\cdot x + y\cdot|\delta(S)| - y\cdot c_S\big)}_{F(S)}

这里用到了 eS=∣S∣−cSe_S = |S| - c_S（树上 SS 的诱导子图是森林，边数 = 点数 − 组件数），cSc_S 是 SS 内部连通分量数。
因此问题化为最大化

F(S)=∣S∣⋅x+y⋅∣δ(S)∣−y⋅cS,F(S)= |S|\cdot x + y\cdot|\delta(S)| - y\cdot c_S,

最后答案再加一个常数 +y+y（当 SS 取遍全体点时，C=0C=0，这个常数会和 −y-y 抵消，仍然正确）。

树形 DP 做法（O(n)）

任选根 rr，对每个点 uu 定义三种状态（值均为“贡献的 FF”而非总 cost）：

  • f0(u)f_0(u)：u∉Su\notin S；
  • f1(u)f_1(u)：u∈Su\in S 且与父亲同属一个 SS 内部连通块（不新开块）；
  • f2(u)f_2(u)：u∈Su\in S 且其父 ∉S\notin S（在 uu 处新开一个 SS 内部连通块，需支付 −y-y 的“开块惩罚”）。

转移（孩子集合为 ch(u)\text{ch}(u)）：

  • 若 u∉Su\notin S：

    f0(u)=∑v∈ch(u)max⁡(  f0(v),  f2(v)+y  )f_0(u)= \sum_{v\in \text{ch}(u)} \max\Big( \;f_0(v),\; f_2(v)+y\;\Big)

    解释：孩子 vv 若进 SS，由于父不在 SS，必须从 vv 开新块（f2f_2 含 −y-y），同时边 (u,v)(u,v) 被切，∣δ∣|\delta| 增 1 给 +y+y，两项在数值上“抵一部分”，但需分别计入。

  • 若 u∈Su\in S 且与父同块：

    f1(u)=x  +  ∑v∈ch(u)max⁡(  f1(v),  f0(v)+y  )f_1(u)= x \;+\; \sum_{v\in \text{ch}(u)} \max\Big( \;f_1(v),\; f_0(v)+y\;\Big)

    解释：uu 取入给 +x+x。孩子若不取入，边被切给 +y+y；若取入，则必须与 uu 相连同块（选 f1(v)f_1(v)）。

  • 若 u∈Su\in S 且父不在 SS：

    f2(u)=x−y  +  ∑v∈ch(u)max⁡(  f1(v),  f0(v)+y  )f_2(u)= x - y \;+\; \sum_{v\in \text{ch}(u)} \max\Big( \;f_1(v),\; f_0(v)+y\;\Big)

    解释：在 uu 新开一个 SS 内部块，先付 −y-y，点费 +x+x，孩子转移与 f1f_1 相同。

根没有父节点，因此最终

F∗=max⁡(f0(r),  f2(r)),答案=y+F∗.F^*=\max\big(f_0(r),\; f_2(r)\big), \qquad \text{答案} = y + F^*.

边界检查

  • 取 S=∅S=\varnothing：DP 会得到 F∗=0F^*=0，答案 =y=y（一次整块爆破整个树）。
  • 取 S=VS=V：DP 会得到 F∗=n⋅x−yF^*=n\cdot x - y，答案 =n⋅x=n\cdot x（全用单点爆破）。

伪代码

text
build tree, root at 1
dfs(u, parent):
  f0 = 0
  f1 = x
  f2 = x - y
  for v in children(u):
    dfs(v, u)
    f0 += max( f0[v], f2[v] + y )
    f1 += max( f1[v], f0[v] + y )
    f2 += max( f1[v], f0[v] + y )
  save f0[u], f1[u], f2[u]

dfs(root, 0)
answer = y + max( f0[root], f2[root] )
print(answer)

时间复杂度 O(n)O(n)，空间 O(n)O(n)。
这套转移不需要特判 xx 与 yy 的大小关系，统一适用，自动给出“全单点”“全整块”以及“两者混合（先切点再整块）”中的最优策略。

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