(树形DP) zoj 3201

Tree of Tree

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 1
10 20 30
0 1
0 2
3 2
10 20 30
0 1
0 2

Sample Output

30
40

题意：选K个节点的子树的最大值

dp[i][j]以i为父亲节点的选j个节点的最大值

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;
int n,k,dp[105][105],val[105];
vector<int> e[105];
void dfs(int u,int father)
{
    dp[u][1]=val[u];
    for(int i=0;i<e[u].size();i++)
    {
        int v=e[u][i];
        if(v==father)
            continue;
        dfs(v,u);
        for(int j=k;j>=1;j--)
        {
            for(int t=1;t<=j;t++)
                dp[u][j]=max(dp[u][j],dp[u][t]+dp[v][j-t]);
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            e[i].clear();
            for(int j=0;j<n;j++)
                dp[i][j]=0;
        }
        for(int i=0;i<n;i++)
            scanf("%d",&val[i]);
        for(int i=1;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            e[x].push_back(y);
            e[y].push_back(x);
        }
        dfs(0,-1);
        int ans=0;
        for(int i=0;i<n;i++)
            ans=max(ans,dp[i][k]);
        printf("%d\n",ans);
    }
    return 0;
}