poj 2342 Anniversary party （树形DP）

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

题意：话说一个公司的一些人要去参加一个party，每个人有一个愉悦值，而如果某个人的直接上司在场的话会非常扫兴，所以避免这样的安排，问给出n个人，每个人的愉悦值以及他们的上司所属关系，问你让那些人去可以让总的愉悦值最大，并求出这个值。
思路：思路：用dp数据来记录价值，开数组用下标记录去或者不去、
则状态转移方程为：

DP[i][1] += DP[j][0],

DP[i][0] += max{DP[j][0],DP[j][1]};其中j为i的孩子节点。

这样，从根节点r进行搜索，最后结果为max{DP[r][0],DP[r][1]}。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int dp[6005][2],vis[6005],a[6005];
void dfs(int root)
{
    vis[root]=1;
    for(int i=1;i<=n;i++)
        if(!vis[i]&&a[i]==root)
        {
            dfs(i);
            dp[root][0]+=max(dp[i][0],dp[i][1]);
            dp[root][1]+=dp[i][0];
        }
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof(dp));
        memset(vis,0,sizeof(vis));
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
           scanf("%d",&dp[i][1]);
        int l,r;
        int root=0;
        while(scanf("%d%d",&l,&r),l+r>0)
        {
            a[l]=r;
            root=r;
        }
        dfs(root);
        printf("%d\n",max(dp[root][0],dp[root][1]));
    }
}