HDUOJ--Anniversary party（树形dp）

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29649    Accepted Submission(s): 10053

 

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

 

 

Source

Ural State University Internal Contest October'2000 Students Session

 

 

#include<bits/stdc++.h>
using namespace std;
const int N=6010;

int dp[N][2];//dp[i][1]表示选择此点，dp[i][0]表示不选择此点
int n;
int v[N];
int a,b;

vector<int>tree[N];//树 

int dfs(int u)
{
	//初赋值 
	dp[u][0]=0;
	dp[u][1]=v[u];
	
	for(int i=0;i<tree[u].size();i++)
	{
		int son=tree[u][i];
		dfs(son);
		
		dp[u][0]+=max(dp[son][1],dp[son][0]);
		dp[u][1]+=dp[son][0];
	}
}
 
int main()
{
	while(cin>>n)
	{

		int father[N];
		
		
		for(int i=1;i<=n;i++)
		{
			cin>>v[i];//相应员工的欢乐程度
			father[i]=-1;//初始化为-1.方便之后查找根节点。 
			tree[i].clear();
		}
		while(cin>>a>>b,a!=0&&b!=0)
		{
			father[a]=b;//a的父亲是b  
			tree[b].push_back(a);//用邻接表建树 
		}
		int t=1;
		while(father[t]!=-1)
		{
			t=father[t];
		}
		
		dfs(t);//从根节点开始深搜
		
		printf("%d\n",max(dp[t][1],dp[t][0])); 
	}
	return 0;
	
}