hdu1520-Anniversary party

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

  AC:

  用树形dp解决

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
int dp[6001][2],w[6001],f[6001];
vector<int> son[6001];
int n,a,b;

void dfs(int x)
{
    for(int i=0;i<son[x].size();i++)
    {
        int tmp=son[x][i];//孩子
        dfs(tmp);         //搜索孩子
        dp[x][0]+=max(dp[tmp][1],dp[tmp][0]);
        dp[x][1]+=dp[tmp][0];   //上司不可以和下属在一起,要间隔开
    }
}


int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            son[i].clear();
            scanf("%d",&w[i]);
            dp[i][0]=0;       //未选取的权值
            dp[i][1]=w[i];    //选取时的权值
            f[i]=-1;          //没有上司
        }
        while(scanf("%d%d",&a,&b),(a+b))
        {
            f[a]=b;              //b是a的上司
            son[b].push_back(a); //将b的下属装入
        }
        for(int i=1;i<=n;i++)
        {
            if(f[i]==-1)         //没有父节点,为根结点,就开始搜索
            {
                dfs(i);
                printf("%d\n",max(dp[i][0],dp[i][1]));
                break;
            }
        }
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/wangtao971115/p/10358319.html