Anniversary party

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 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

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

The question ： Yes n personal , Next n The row corresponds to everyone's happiness score , Yes n Group correspondence , If someone goes , Then its leaders cannot go , If this person doesn't go , Its leaders can go or not , On the contrary, the same is true , Ask how to arrange the highest happiness score to output the highest score .

Ideas ： The question is dp problem ,dp[i][0] It means the first one i I will not go ,dp[i][1] It means the first one i Go alone ., If the boss boss Go to the dance ,i I have no choice but not to go , If the boss boss If not ,i You don't have to go , So the state transfer equation is ：

dp[boss][1]=dp[boss][1]+dp[i][0];

dp[boss][0]=dp[boss][0]+max(dp[i][0],dp[i][1])

Finally, compare the maximum output of that value , The code is as follows ：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n;                            //dp[i][0]  The first i If you don't go, you can get the maximum ,dp[i][1] The first i The maximum value that a person can go to 
int f[6020],dp[6020][2],book[6020];
void dfs(int boss)
{
    book[boss]=1;      // This node has been used by the boss 
    for(int i=1;i<=n;i++)
    {
        if(!book[i]&&f[i]==boss) // look for boss Subordinates of 
        {
            dfs(i); // Recursion by subordinates 
            dp[boss][1]=dp[boss][1]+dp[i][0];  // If the boss boss Go to the dance ,i I have no choice but not to go 
            dp[boss][0]=dp[boss][0]+max(dp[i][0],dp[i][1]); // If the boss boss If not ,i You don't have to go , So take the maximum 
        }
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        int i,x,y,boss=0;
        memset(book,0,sizeof(book));
        memset(dp,0,sizeof(dp));
        memset(f,0,sizeof(f));
        for(i=1; i<=n; i++)
            scanf("%d",&dp[i][1]);// Everyone's happiness when everyone goes 
        while(~scanf("%d%d",&x,&y)&&(x||y))
        {
            f[x]=y; //x My boss is y
            boss=y;  // Record any boss 
        }
        dfs(boss); // Find the maximum value by recursion 
        printf("%d\n",max(dp[boss][0],dp[boss][1]));// from boss Start recursion 
    }
    return 0;
}