poj 2342 树形dp(入门题)

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

题意：公司举办party,邀请员工参加，但是每个员工都不喜欢和自己的直接上司一起参加，所以，若上司去了，他的直系下属不能去，反之可以。每个员工都有一个兴奋值，公司想要举办的party的兴奋值尽可能的大。

思路：上网查的博客，初次接触树形dp,没有形成树形dp的思维，此题的方法是用递归写的，较好理解，也是树形dp的入门题

 比较容易理解。

 dp[i][j],i表示节点，j标记状态,j=1时，表示i去参加，j=0时 表示不去参加 。

AC代码：

#include 
#include 
#include 
using namespace std;
int father[6100];
int dp[6100][2];
int book[6100];
int num[6100];
int n,m; 
void dfs(int node)
{
	book[node]=1;
	for(int i=1;i<=n;i++)
	{
		if(!book[i]&&father[i]==node)
		{
			dfs(i);
			dp[node][1]+=dp[i][0];  //这个是上司要去，下属不去 
			dp[node][0]+=max(dp[i][1],dp[i][0]);
			// 这个是上司不去，下属可能去也可能不去 
		}
	}
}
int main()
{
	while(~scanf("%d",&n))
	{
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=n;i++)
		scanf("%d",&dp[i][1]);
		
		for(int i=1;i<=n;i++)
        father[i]=i;
        
		int x,y,root;
		while(scanf("%d %d",&x,&y)&&(x+y))
			father[x]=y;
		for(int i=1;i<=n;i++)
		if(father[i]==i)
		{
		  root=i;
		  break;
		}
		memset(book,0,sizeof(book));
	//	printf("%d\n",root);
		dfs(root);
		int term=dp[root][1]>dp[root][0]?dp[root][1]:dp[root][0];
		
		printf("%d\n",term);
	}
	return 0;
}