hdu 1520 Anniversary party(树形dp)

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8386    Accepted Submission(s): 3641

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

 

题目大意：

        有一个party，希望邀请一个公司内的n个人，这n个人中，有的人有一个直接的上司，而有直接上司的人不想和上司同时参加这个party。每个人有一个愉快度，怎样选择可以让参加party的人的愉快度和最大。

解题思路：

        简单树形dp，dp[i][0]表示第i个人不来时他的下属（直接下属，间接下属）可以获得的最大愉快度，dp[i][1]表示第i个人来时他和他的下属（直接下属，间接下属）可以获得的最大愉快度。

代码：

#include <iostream>
#include <vector>
#include <stdio.h>
#include <cstring>

using namespace std;

int n;
int father[6005],dp[6005][2],v[6005];
vector<int> son[6005];

int dfs(int x){
    if(dp[x][1]) return max(dp[x][1],dp[x][0]);
    dp[x][1] = v[x];
    for(int i = 0; i < son[x].size(); i ++){
        dfs(son[x][i]);
        dp[x][0] += max(dp[son[x][i]][0],dp[son[x][i]][1]);
        dp[x][1] += dp[son[x][i]][0];
    }
    return max(dp[x][1],dp[x][0]);
}

int main()
{
    int l,k;
    while(scanf("%d",&n) != EOF){
        memset(father,0,sizeof(father));
        memset(son,0,sizeof(son));
        memset(dp,0,sizeof(dp));
        for(int i = 1; i <= n; i ++) son[i].clear();
        for(int i = 1; i <= n; i ++){
            scanf("%d",&v[i]);
        }
        while(true){
            scanf("%d%d",&l,&k);
            if(l == 0) break;
            father[l] = k;
            son[k].push_back(l);
        }
        int ans = 0;
        for(int i = 1; i <= n; i ++){
            if(father[i] == 0) ans += dfs(i);
        }
        printf("%d\n",ans);
    }
    return 0;
}