hdu 1520Anniversary party(简单树形dp)

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4310    Accepted Submission(s): 1976

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

 

Source

Ural State University Internal Contest October'2000 Students Session

 

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题目大意：给你一颗树，是人际关系树，然后互为上下级的（父子节点）的不能同一时候选中。

解题思路：树形dp，任意从一个点開始扩展，把周围全部节点的dp都解决出来，然后加上去就可以。

用dp[i][0]表示不选中i号，周围的都能够选的最大值，dp[i][1]表示选中i号，那么周围都不能选的最大值。

则

dp[i][0]+=(dp[j][1],dp[j][0])

dp[i][1]+=dp[j][0]

i和j相邻，而且在求i的时候从j扩展的都已经求出来。

详见代码：

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1520

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
using namespace std;
const int maxn=6005;

vector<int> mq[maxn];
int dp[maxn][2];

void dfs(int cur,int p)
{
    int i,nex;
    for(i=0;i<mq[cur].size();i++)
    {
        nex=mq[cur][i];
        if(nex==p) continue;

        dfs(nex,cur);  //先找从cur这个点能够扩展的dp
        dp[cur][0]+=max(dp[nex][0],dp[nex][1]);
        dp[cur][1]+=dp[nex][0];
    }
}

int main()
{
    int n,i;
    int a,b;

    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            mq[i].clear();
            scanf("%d",&dp[i][1]);
        }

        while(scanf("%d%d",&a,&b)&&(a+b))
        {
            mq[a].push_back(b);
            mq[b].push_back(a);
        }

        dfs(1,0);

        int ans=max(dp[1][0],dp[1][1]);
        printf("%d\n",ans);
    }

    return 0;
}