POJ2342 Anniversary party

最新推荐文章于 2019-08-08 13:46:36 发布

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最新推荐文章于 2019-08-08 13:46:36 发布

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原文链接：http://www.cnblogs.com/zitian246/p/9123593.html

本文介绍了一道经典的动态规划题目——Ural大学80周年庆典派对问题，通过构建员工间的层级关系避免邀请上下级同时参加，以达到嘉宾兴奋值总和的最大化。

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链接：http://poj.org/problem?id=2342

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10326		Accepted: 5913

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

Sample Output

Source

Ural State University Internal Contest October'2000 Students Session

树形DP：看着像搜索哈。。。

题意：公司人要开party，要从N个人之间选几人是兴奋值最大（2--N+1行，每个人都有唯一的兴奋值）

剩下几行是从属关系，左边的是右边的直接下属，但选择的时候，要避开这种直接领导关系，最多只能去一人，问兴奋值最大是多少？

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define maxn 7000
using namespace std;
int dp[maxn][2];//1 去 0 不去
int father[maxn];
int vis[maxn];
int n;
void dfs(int root)
{
    for(int i=1;i<=n;i++)
    {
        if(father[i]==root)
        {
            dfs(i);
            dp[root][1]+=dp[i][0];
            dp[root][0]+=max(dp[i][0],dp[i][1]);
        }
    }
}
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&dp[i][1]);
        }
        int l,k;
        int root=0;
        int flag=1;
        memset(vis,0,sizeof(vis));
        while(scanf("%d%d",&l,&k),l||k)
        {
            father[l]=k;
            if(root==l||flag)
            {
                root=k;
            }
        }
        while(father[root])
        {
            root=father[root];
        }
        dfs(root);
        int ans=max(dp[root][0],dp[root][1]);
        printf("%d\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/zitian246/p/9123593.html