ZOJ 3201 Tree

Tree of Tree

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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

3 110 20 300 10 23 210 20 300 10 2

Sample Output

3040

题目链接http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3326

题意：

从一颗 n 个节点的树中选出含有 k 个节点的子树，求子树最大权值

分析：

树形背包dp 

dp[ fa ][ x ] 表示以 fa 为根节点，选 x 个节点的最大权值

则状态转移方程为：dp[ fa ][ fax ] = max(dp[ fa ][ fax ], dp[ fa ][ fax - sonx ] + dp[ son ][ sonx ])( 1 < sonx < fax )

对于 fa 的每个孩子，可以选择 1,2,3...fax - 1个节点分配给它

代码：

#include <cstdio>#include <cstring>#include <vector>#include <iostream>#include <algorithm>using namespace std;const int MAXN = 100 + 5;int n, k;vector<int> adj[MAXN];              //邻接表，adj[x] 中保存所有与 x 相连的节点int dp[MAXN][MAXN], vis[MAXN], weight[MAXN];void dfs(int fa){    dp[fa][1] = weight[fa];         //以fa为根，只有一个节点的话结果就是它本身    vis[fa] = 1;                    //标记为已经访问    int len = adj[fa].size();       //得到与它相连的节点的个数    for (int i = 0; i < len; i++)    {        int son = adj[fa][i];       //得到子节点编号        if (!vis[son])              //如果还未访问        {            dfs(son);               //先递归子节点求值            for (int fax = k; fax > 0; fax--)           //状态转移方程                for (int sonx = 1; sonx < fax; sonx++)                    dp[fa][fax] = max(dp[fa][fax], dp[fa][fax - sonx] + dp[son][sonx]);        }    }}int main(){    while (~scanf("%d%d", &n, &k))    {        for (int i = 0; i < n; i++)        {            scanf("%d", &weight[i]);            adj[i].clear();             //先将邻接表清空        }        for (int i = 1; i < n; i++)        {            int x, y;            scanf("%d%d", &x, &y);            adj[x].push_back(y);        //加入各自的表中            adj[y].push_back(x);        }        memset(dp, 0, sizeof(dp));        memset(vis, 0, sizeof(vis));        dfs(0);                         //从节点 0 开始深搜        int ans = 0;        for (int i = 0; i < n; i++)            ans = max(ans, dp[i][k]);        printf("%d\n", ans);    }    return 0;}