HDU 1520 Anniversary party（DFS或树形DP）

Problem Description

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

题意：n个人去參加聚会。当中有直接上下级关系的參加聚会会影响气氛，所以不能參加。每一个人參加聚会都会有一个活跃度，如今要你计算从n个人中选择出一些人，使得本次聚会的活跃度最大

思路：非常easy，每一个人參加和不參加都会有不同的结果，他參加，那么仅仅须要加上他的下属不參加的权值，他不參加。那么加上他下属參加或不參加中的最大值，最后仅仅须要计算出最高上司參加或者不參加的最大值就能够了

PS：树形DP的代码你们去參考其它人的吧，近期在训练蓝桥杯，所以强化DFS能力

AC代码：

#include<cstdio>

#define N 6005

struct p
{
    int fm;
    int child;
    int brother;
    int att;
    int no;
    void init()
    {
        no=0;
        fm=0;
        brother=0;
        child=0;
    }
    int Max()
    {
        return att>no? att:no;
    }
}num[N];

void dfs(int x)
{
    int child=num[x].child;
    while(child)
    {
        dfs(child);
        num[x].att+=num[child].no;
        num[x].no+=num[child].Max();
        child=num[child].brother;
    }
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            num[i].init();
            scanf("%d",&num[i].att);
        }
        int a,b;
        while(scanf("%d %d",&a,&b),a||b)
        {
            num[a].fm=b;
            num[a].brother=num[b].child;
            num[b].child=a;
        }
        for(int i=1;i<=n;i++)
        {
            if(num[i].fm==0)
            {
                dfs(i);
                printf("%d\n",num[i].Max());
                break;
            }
        }
    }
    return 0;
}