HDu 1520 Anniversary party(ベースツリーDP)

1、http://acm.hdu.edu.cn/showproblem.php?pid=1520
2、テーマ大意:
n人の従業員がいて、すべての従業員はすべて1つのrating値があって、従業員の間の上下関係を与えて、直接上下関係の従業員が同時に出席することができないことを要求して、今要求するのはどの従業員が出席することを選んで、彼らのratingの和を最大にすることができます
定義dp[i][1]はi従業員の出席時の最大値を表し、dp[i][0]はi従業員の出席しない時の最大値を表す
dp[i][1]=dp[v][0]//i社員が出席すると、彼のすべての直接部下vが出席しないときの最大値に等しい
dp[i][0]=max(dp[v][1],dp[v][0])/i社員が出席しない場合、部下が出席または出席しない最大値に等しい
3、テーマ:
Anniversary party
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3617    Accepted Submission(s): 1680
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
   
   
   
   

 
4、ACコード:

#include<stdio.h>
#include<vector>
using namespace std;
#define N 6005
vector<int> vec[N];
int rating[N];
int f[N];
int dp[N][2];//dp[i][1]  i    ，dp[i][0]  i     
void dfs(int root)
{
    dp[root][1]=rating[root];
    for(int i=0; i<vec[root].size(); i++)
    {
        int v=vec[root][i];
        dfs(v);
        dp[root][1]+=dp[v][0];
        dp[root][0]+=max(dp[v][1],dp[v][0]);
    }
}
int main()
{
    int n,l,k;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&rating[i]);
            f[i]=-1;//     
            vec[i].clear();
            dp[i][0]=0;
            dp[i][1]=0;
        }
        while(scanf("%d%d",&l,&k)!=EOF)
        {
            if(l==0 && k==0)
                break;
            vec[k].push_back(l);
            f[l]=k;
        }
        int a=1;
        while(f[a]!=-1)
        {
            a=f[a];
        }
        dfs(a);
        printf("%d
",max(dp[a][1],dp[a][0]));
    }
    return 0;
}