[HDU 1561] The more, The Better 树形dp

http://acm.hdu.edu.cn/showproblem.php?pid=1561

题意：中文题。

思路：树形dp，转移方程：
dp[i][k] = max(dp[i][k], dp[i][k-l]+dp[son][l]);

#include <cstdio>#include <vector>#include <cstring>#include <iostream>using namespace std;int n, m;int val[205];int dp[205][205];vector<int>eg[205];int Dfs(int rt, int pre){    for(int i = 1; i <= m; i++){        dp[rt][i] = val[rt];    }    int len = eg[rt].size();    for(int i = 0; i < len; i++){        int son = eg[rt][i];        if(son == pre)            continue;        Dfs(son, rt);        for(int k = m; k > (rt != 0); k--) //因为 0 虚拟根，不需要选        {            for(int l = 1; l + (rt != 0) <= k; l++)            {                if(dp[rt][k] < dp[rt][k-l] + dp[son][l]){                    dp[rt][k] = dp[rt][k-l] + dp[son][l];                }            }        }    }    return 0;}int main(){    val[0] = 0;    while(cin>>n>>m && (m || n)){        int x, y;        memset(dp, 0, sizeof(dp));        for(int i = 0; i <= n; i++){            eg[i].clear();        }        for(int i = 1; i <= n; i++){            cin>>x>>val[i];            eg[x].push_back(i);        }        if(m == 0){            cout<<"0"<<endl;            continue;        }        Dfs(0, -1);        cout<<dp[0][m]<<endl;    }    return 0;}