ZOJ3201(树形DP)

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

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

Sample Output

30
40
题意：求大小为k权值最大的子树。 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN=105;
vector<int> tree[MAXN];
int n,k;
int w[MAXN];
int res;
int dp[MAXN][MAXN];
void dfs(int u,int fa)
{
    dp[u][1]=w[u];
    for(int i=0;i<tree[u].size();i++)
    {
        int v=tree[u][i];
        if(v==fa)
            continue;
        dfs(v,u);
        for(int j=k;j>=0;j--)
            for(int l=1;l<=j;l++)
                dp[u][k]=max(dp[u][k],dp[u][l]+dp[v][j-l]);
    }
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        res=0;
        for(int i=0;i<n;i++)
        {
            tree[i].clear();
            scanf("%d",&w[i]);
        }
        for(int i=0;i<n-1;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            tree[u].push_back(v);
            tree[v].push_back(u);
        }
        dfs(0,-1);
        for(int i=0;i<n;i++)
            res=max(dp[i][k],res);
        printf("%d\n",res);
        
    }
    
    return 0;
}