poj 2342 Anniversary party （树形DP）

地址 http://poj.org/problem?id=2342

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8096		Accepted: 4631

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

Source

Ural State University Internal Contest October'2000 Students Session

 

分析：树形DP的模版题

首先要找到根节点，也就是职位最高的人,然后从根节点进行DP

dp[x][1]代表x到场的，以x为根节点的总体的愉悦度。

dp[x][0]代表x未到场

状态转移方程

do[x][1]+=dp[i][0]  其中x为i的上司，由题意故i不能到场

 dp[x][0]+=max(dp[i][1],dp[i][0]);

 

代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int dp[6200][2];
int pre[6200];
int vis[6200];
    int n,root,a,b,maxx;
void dfs(int x)
{
    vis[x]=1;
    for(int i=1;i<=n;i++)
    {
        if(pre[i]==x&&!vis[i])
        {
            dfs(i);
            dp[x][1]+=dp[i][0];
            dp[x][0]+=max(dp[i][1],dp[i][0]);
        }
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        memset(dp,0,sizeof(dp));
        memset(pre,0,sizeof(pre));
        for(int i=1;i<=n;i++)
        {
        pre[i]=i;
        scanf("%d",&dp[i][1]);
        }
        while(scanf("%d%d",&a,&b)!=EOF)
        {
            if(a==0&&b==0) break;
            pre[a]=b;
        }
        root=1;
        while(pre[root]!=root)
          root=pre[root];
          dfs(root);
          maxx=max(dp[root][1],dp[root][0]);
          printf("%d
",maxx);
    }
    return 0;
}