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 KIt means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line0 0
Output
Output should contain the maximal sum of guests’ ratings.
Sample Input
711111111 32 36 47 44 53 50 0
Sample Output
5
题意：n个人去参加聚会，其中有直接上下级关系的参加聚会会影响气氛，所以不能参加，每个人参加聚会都会有一个活跃度，现在要你计算从n个人中选择出一些人，使得本次聚会的活跃度最大
思路：很简单，每个人参加和不参加都会有不同的结果，他参加，那么只需要加上他的下属不参加的权值，他不参加，那么加上他下属参加或不参加中的最大值，最后只需要计算出最高上司参加或者不参加的最大值就可以了
PS：树形DP的代码你们去参考其他人的吧，，最近在训练蓝桥杯，所以强化DFS能力
AC代码：
#include<cstdio>#define N 6005struct 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;}
生命不息，在任何一种博大的辉煌之后，都掩藏着许多鲜为人知的艰难的奋斗。
