F - Anniversary party

HDU - 1520

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：一个公司的领导结构成树形结构，每个人都有一个领导，现在要开一个party，想邀请互相不是上下级关系的人来参加，每个人都有一个参加party的开心度，怎么选人使开心值最大，输出最大开心值

解法：树形dp，对于每个结点有dp[i][0]为此人不参加的最大快乐值，dp[i][1]为此人参加的最大快乐值，dp[i][1]+=dp[i-1][1]，dp[i][0]=max(dp[i-1][0],dp[i-1][1])

代码：

#include<iostream>
#include<vector>
using namespace std;
int dp[6005][2];
bool root[6005];
vector<int> a[6005];
void dfs(int b){
    if(a[b].empty())
        return;
    for(int i=0;i<a[b].size();i++){
        dfs(a[b][i]);
        dp[b][1]+=dp[a[b][i]][0];
        dp[b][0]+=max(dp[a[b][i]][0],dp[a[b][i]][1]);
    }
    return;
}
int main()
{
    int n,x,y,r;
    while(cin>>n){
        for(int i=1;i<=n;i++){
            cin>>dp[i][1];
            dp[i][0]=0;
            root[i]=1;
            a[i].clear();
        }
        while(cin>>x>>y&&x&&y){
            a[y].push_back(x);
            root[x]=0;
        }
        for(int i=1;i<=n;i++){
            if(root[i]==1){
                r=i;
                break;
            }
        }
        dfs(r);
        cout<<max(dp[r][0],dp[r][1])<<endl;
    }
    return 0;
}