hdu 1520 Anniversary party(基本树形DP)

1、http://acm.hdu.edu.cn/showproblem.php?pid=1520

2、题目大意：

有n个员工，每个员工都有一个rating值，给出员工之间的上下级的关系，要求是有直接上下级关系的员工不能同时出席，现在要求的是选择哪些员工出席，会使得他们的rating之和最大

定义dp[i][1]表示i员工出席时的最大值，dp[i][0]表示i员工不出席时的最大值

dp[i][1]=dp[v][0]//当i员工出席时，就等于他的所有直接下属v不出席时的最大值

dp[i][0]=max(dp[v][1],dp[v][0])//当i员工不出席时，就等于他的下属出席或者不出席的最大值

3、题目：

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代码：

#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]表示i结点选择，dp[i][0]表示i结点不选择
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;//标记根节点
            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\n",max(dp[a][1],dp[a][0]));
    }
    return 0;
}