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 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 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

这题和poj2342是一样的，但是hdu的数据加强了，用poj2342AC的代码可能会超时，poj2342解析报告

这里用的方法有点不一样，但大概也都是一样的，只是加入了一些父子兄弟关系，也不是很难理解。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))

struct Tree
{
    int father;//父亲
    int child;//儿子
    int brother;//兄弟
    int Not;//不去party
    int TakeParty;//去party
    int MAX() {return TakeParty > Not ? TakeParty : Not;}//结构体调用函数省时间
    int init()//树清空
    {
        father = child = brother = Not = 0;
    }
}tree[6005];

void dfs(int idx)
{
    int child = tree[idx].child;
    while(child)//如果有孩子继续查
    {
        dfs(child);
        tree[idx].TakeParty += tree[child].Not;//如果idx去party，那么它的孩子就不能去
        tree[idx].Not += tree[child].MAX();//如果idx不去party，那么它的孩子可去可不去
        child = tree[child].brother;//查找其他孩子
    }
}

int main()
{
    int n,a,b;
    while(scanf("%d",&n)==1)
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&tree[i].TakeParty);
            tree[i].init();
        }
        while(scanf("%d%d",&a,&b),a+b)//建树
        {
            tree[a].father = b;
            tree[a].brother = tree[b].child;
            tree[b].child = a;
        }
        for(int i=1; i<=n; i++)//查找
        {
            if(!tree[i].father)
            {
                dfs(i);
                printf("%d\n",tree[i].MAX());
                break;
            }
        }
    }
    return 0;
}