ZJU 3201 Tree of Tree 树状DP

最新推荐文章于 2020-03-08 19:50:52 发布

HQD因为有趣所以做题

最新推荐文章于 2020-03-08 19:50:52 发布

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分类专栏： ACM_动态规划文章标签： tree integer each graph output input

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探讨了在一个带有节点权重的树形结构中寻找指定大小的最大子树的问题。通过深度优先搜索(DFS)算法实现解决方案，并详细展示了核心代码片段。

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Tree of Tree

Time Limit: 1 Second Memory Limit: 32768 KB

You're given a tree with weights of each node, you need to find the maximum subtree of specified size of this tree.

Tree Definition
A tree is a connected graph which contains no cycles.

Input

There are several test cases in the input.

The first line of each case are two integers N(1 <= N <= 100), K(1 <= K <= N), where N is the number of nodes of this tree, and K is the subtree's size, followed by a line with N nonnegative integers, where the k-th integer indicates the weight of k-th node. The following N - 1 lines describe the tree, each line are two integers which means there is an edge between these two nodes. All indices above are zero-base and it is guaranteed that the description of the tree is correct.

Output

One line with a single integer for each case, which is the total weights of the maximum subtree.

Sample Input

Sample Output

30
40

Author: LIU, Yaoting

这个题目是无向图，而且是任何一个点都可能是根节点，这个程序算出来的任何F[I][K]都是以I为根节点的最大值，呵呵

关键代码

void dfs(int idx)

{

    for(int i=bug[idx]; i<=m; i++)

        dp[idx][i]=p[idx];//初始化数组

    used[idx]=1;//标记，以便剪枝

    for(int i=0; i<num[idx]; i++)

    {

        if(used[map[idx][i]]) continue;

        dfs(map[idx][i]);//直到搜索到叶子节点

        for(int j=m; j>=bug[idx]; j--)

            for(int k=1; k<=j-bug[idx]; k++)

                if(dp[idx][j]<dp[idx][j-k]+dp[map[idx][i]][k])

                    dp[idx][j]=dp[idx][j-k]+dp[map[idx][i]][k];

    }

}