HDU 1520 Anniversity party

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. 

InputEmployees 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 0OutputOutput 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
题解：树形DP。对于每个人，我们储存下比他职位低的下标，并对比他职位低的人打上标记代表这个人有上司。dp[i][1]代表选i这个人，dp[i][0]代表不选这个人，然后从底层开始，对上一层dp[x][1]+=dp[x-1][0];
dp[x][0]+=max(dp[x-1][1],dp[x-1][0]);
然后最大值为： max(dp[root][1],dp[root][0]);root为没有标记的那个人，即没有任何下属的人；
参考代码为：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cstdlib>
#include<set>
using namespace std;
const int maxn=1e4+10;
int N,u,v,w[maxn],vis[maxn],dp[maxn][2];
vector<int> vec[maxn];

void solve(int x)
{
	for(int i=0;i<vec[x].size();i++)
	{
		solve(vec[x][i]);
		dp[x][1]+=dp[vec[x][i]][0];
		dp[x][0]+=max(dp[vec[x][i]][0],dp[vec[x][i]][1]);
	}
}

int main()
{
	while(~scanf("%d",&N))
	{
		for(int i=0;i<=N;i++) vec[i].clear();
		memset(vis,0,sizeof vis);
		memset(dp,0,sizeof dp);
		for(int i=1;i<=N;i++) scanf("%d",w+i);
		for(int i=1;i<=N;i++) dp[i][1]=w[i];
		while(scanf("%d%d",&u,&v), u+v)
		{
			vec[v].push_back(u);
			vis[u]=1;
		}
		int flag;
		for(int i=1;i<=N;i++)
		{
			if(!vis[i])
			{
				solve(i);
				flag=i;
				break;
			}
		}
		int Max=max(dp[flag][1],dp[flag][0]);
		printf("%d\n",Max);
	}
	return 0;
}