HDU 1520 Anniversary party（DFS或樹形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

題意：n個人去參加聚會，其中有直接上下級關系的參加聚會會影響氣氛，所以不能參加，每個人參加聚會都會有一個活躍度，現在要你計算從n個人中選擇出一些人，使得本次聚會的活躍度最大
思路：很簡單，每個人參加和不參加都會有不同的結果，他參加，那麼只需要加上他的下屬不參加的權值，他不參加，那麼加上他下屬參加或不參加中的最大值，最後只需要計算出最高上司參加或者不參加的最大值就可以了
PS：樹形DP的代碼你們去參考其他人的吧，最近在訓練藍橋杯，所以強化DFS能力
AC代碼：

#include

#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;
}