Anniversary party （树形DP）

There is going to be a party to celebrate the 80-th Anniversary of the Ural State University. The University has a hierarchical structure of employees. It means that the supervisor relation forms a tree rooted at the rector V. E. Tretyakov. In order to make the party funny for every one, the rector does not want both an employee and his or her immediate supervisor to be present. The personnel office has evaluated conviviality of each employee, so everyone has some number (rating) attached to him or her. Your task is to make a list of guests with the maximal possible sum of guests' conviviality ratings. 

Input

Employees are numbered from 1 to N. A first line of input contains a number N. 1 <= N <= 6 000. Each of the subsequent N lines contains the conviviality rating of the corresponding employee. Conviviality rating is an integer number in a range from -128 to 127. After that go T lines that describe a supervisor relation tree. Each line of the tree specification has the form: 
L K 
It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line 
0 0

Output

Output should contain the maximal sum of guests' ratings. 
Sample Input

7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0

Sample Output

5
题目大意：
先输入n，代表n个人，下面n行代表n个人的权值，在输入若干对数字l，k,代表k是l的上司，现在有一个晚会，n个人中若干个人参加，要求参加的人不存在上司与下属的关系，
求参加的人最大的权值和。

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;
int in[6005],dp[6005][2],n;
vector<int> vec[6005];
void dfs(int root)
{
    for(int i=0;i<vec[root].size();i++)
    {
        int x=vec[root][i];
        dfs(x);
        dp[root][1]+=dp[x][0];///上司参加，下属一定不参加
        dp[root][0]+=max(dp[x][0],dp[x][1]);///上司不参加，加上下属参加和不参加的最大值
    }
}
int main()
{
    while(cin>>n)
    {
        memset(in,0,sizeof in);
        memset(dp,0,sizeof dp);
        for(int i=1;i<=n;i++)
            cin>>dp[i][1],vec[i].clear();
        int x,y;
        while(cin>>x>>y,x||y)
            vec[y].push_back(x),in[x]++;
        for(int i=1;i<=n;i++)
        {
            if(!in[i])
            {
                dfs(i);
                cout<<max(dp[i][1],dp[i][0])<<'
';
                break;
            }
        }
    }
    return 0;
}