Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9128		Accepted: 5250

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.

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 N – 1 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 should contain the maximal sum of guests' ratings.

7

1

1

1

1

1

1

1

1 3

2 3

6 4

7 4

4 5

3 5

0 0

5

 #include<iostream>

 #include<cmath>

 #include<algorithm>

 #include<vector>

 #include<cstdio>

 #include<cstdlib>

 #include<cstring>

 #include<string>  

 using namespace std;  

 #define maxn 6005  

 int n;

 int dp[maxn][],father[maxn];//dp[i][0]0表示不去，dp[i][1]1表示去了

 bool visited[maxn];  

 void tree_dp(int node)

 {

     int i;

     visited[node] = ;

     for(i=; i<=n; i++) //找1到n中找node节点的下属

     {

         if(!visited[i]&&father[i] == node)//i为下属

         {

             tree_dp(i);//递归调用孩子结点，从叶子结点开始dp

             //关键

             dp[node][] += dp[i][];//上司来,下属不来

             dp[node][] +=max(dp[i][],dp[i][]);//上司不来，下属来、不来

         }

     }

 }  

 int main()

 {

     int i;

     int f,c,root;

     while(scanf("%d",&n)!=EOF)

     {

         memset(dp,,sizeof(dp));

         memset(father,,sizeof(father));

         memset(visited,,sizeof(visited));

         for(i=; i<=n; i++)

         {

             scanf("%d",&dp[i][]);

         }

         root = ;//记录父结点

         bool beg = ;

         while (scanf("%d %d",&c,&f),c||f)

         {

             //c的老爸是f

             father[c] = f;

             //儿子是跟节点   或者这条边是第一条边

             //父亲做根节点

             if( root == c || beg )

             {

                 root = f;

             }

         }

         //循环寻找根节点

         while(father[root])//查找父结点

             root=father[root];

         //从根节点开始树形dp

         tree_dp(root);

         //取最大的上司去或者不去中的大值

         int imax=max(dp[root][],dp[root][]);

         printf("%d\n",imax);

     }

     return ;  

 }