poj 2342 Anniversary party（树形dp）

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6070		Accepted: 3497

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 N – 1 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

Source

Ural State University Internal Contest October'2000 Students Session

任何一个点的取舍可以看作一种决策，那么状态就是在某个点取的时候或者不取的时候，以他为根的子树能有的最大活跃总值。分别可以用f[i,1]和f[i,0]表示第i个人来和不来。

当i来的时候，dp[i][1] += dp[j][0];//j为i的下属

当i不来的时候，dp[i][0] +=max(dp[j][1],dp[j][0]);//j为i的下属

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int inf = 2147483647;
const double PI = acos(-1.0);
const double e = 2.718281828459;
const int mod = 1000000009;
typedef long long LL;
//freopen("in.txt","r",stdin); //输入重定向,输入数据将从in.txt文件中读取
//freopen("out.txt","w",stdout); //输出重定向,输出数据将保存在out.txt文件中cin
int dp[7000][2];
int v[7000], f[7000], vis[7000], n, m;
void dp_dfs(int root)
{
	vis[root] = 1;
	for (int i = 1; i <= n; ++i)
	{
		if (f[i] == root && !vis[i])
		{
			dp_dfs(i);
			dp[root][1] += dp[i][0];
			dp[root][0] += max(dp[i][0], dp[i][1]);
		}		
	}
}
int main()
{
	int  i, j;
	while (~scanf("%d", &n))
	{
		memset(dp, 0, sizeof(dp));
		for (i = 1; i <= n; ++i)
		{
			scanf("%d", &dp[i][1]);
		}
		int L, K;
		memset(f, 0, sizeof(f));
		memset(vis, 0, sizeof(vis));
		int root;
		while (~scanf("%d%d", &L, &K))
		{
			if (L == 0 && K == 0)
				break;
			f[L] = K;
			root = L;
		}
		while (f[root] != 0)
		{
			root = f[root];//因为只存在一颗树， 所以只需找到根节点然后往下递归即可
		}
		dp_dfs(root);
		printf("%d\n", max(dp[root][0], dp[root][1]));
			
	}
}