HDU - 1561 (dp + tree knapsack problem)

Topic: the Click
each castle is overcome, there must be a specific castle was captured before, in order to capture the castle.
Forming a relationship of parent and child nodes, Lenovo to take the tree dp resolved.
Child node must first go down its parent node, a contribution, if not the parent node is 0;
DP [i] [j]: i is represented by a node, it takes a maximum value j obtained castle treasures.

#include<cmath>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<istream>
#include<vector>
#include<stack>
#include<map>
#include<algorithm>
#include<queue>
#define MAX_len 50100*4
using namespace std;
typedef long long ll;
int dp[210][210];// 第i个节点 收集了j个 的最大值
vector<int>hh[210];
int a[210];
int n,m;
bool vis[210];
void dfs(int root)
{
    if(vis[root])
        return ;
    vis[root]=true;
    dp[root][1]=a[root];
    for(int i=0;i<hh[root].size();i++)
    {
        int v=hh[root][i];
        if(vis[v])
            continue;
            dfs(v);
            for(int j=m+1;j>=2;j--)
            {
                for(int k=1;k<j;k++)
                    dp[root][j]=max(dp[root][j],dp[root][j-k]+dp[v][k]);
            }
    }
}
int main()
{
    while(scanf("%d %d",&n,&m)&&(n||m))
    {
        memset(vis,false,sizeof(vis));
        memset(dp,0,sizeof(dp));
        a[0]=0;
        int i,j,k;
        for(i=0;i<=n;i++)
            hh[i].clear();
        for(i=1;i<=n;i++)
        {
            int x;
            scanf("%d %d",&x,&a[i]);
            {
                hh[i].push_back(x);
                hh[x].push_back(i);
            }
        }
        dfs(0);
        printf("%d\n",dp[0][m+1]);
    }
    return 0;
}