HDU-1520 Anniversary party（树形DP）

最新推荐文章于 2025-08-20 15:04:37 发布

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最新推荐文章于 2025-08-20 15:04:37 发布

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文章标签： java

原文链接：http://www.cnblogs.com/dacc123/p/8228852.html

本文探讨如何在乌拉尔国立大学庆祝80周年纪念日的派对上，根据员工的社交评分，制定出既不违反上级与下级不能同时出席的规定，又能最大化邀请员工总评分的策略。通过解析输入数据，包括员工数量、社交评分及上下级关系，最终输出邀请名单的最大可能总评分。

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Anniversary party

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7566 Accepted Submission(s): 3321

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
这道题目和这一道几乎一样的，
http://blog.youkuaiyun.com/dacc123/article/details/50354675
换汤不换药，
看懂了这篇博客的代码，这题自然就会了

#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>

using namespace std;
#define MAX 6000
struct Node
{
    int value;
    int next;
}edge[MAX*2+5];
int tag[MAX+5];
int dp[MAX+5][2];
int n;
int root;
int tot;
int rat[MAX+5];
int head[MAX+5];
int vis[MAX+5];
void add(int x,int y)
{
    edge[tot].value=y;
    edge[tot].next=head[x];
    head[x]=tot++;
}
void dfs(int root)
{

    dp[root][1]=rat[root];
    dp[root][0]=0;
    vis[root]=1;
    for(int i=head[root];i!=-1;i=edge[i].next)
    {
        int k=edge[i].value;
        if(!vis[k])
        {
            dfs(k);
            dp[root][1]+=dp[k][0];
            dp[root][0]+=max(dp[k][0],dp[k][1]);
        }
    }


}
int main()
{
    int a,b;
    while(scanf("%d",&n)!=EOF)
    {
        tot=0;
        memset(head,-1,sizeof(head));
        memset(dp,0,sizeof(dp));
        memset(vis,0,sizeof(vis));
        memset(tag,0,sizeof(tag));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&rat[i]);
        }
        scanf("%d%d",&a,&b);
        while(a!=0&&b!=0)
        {
            tag[a]=1;
            add(b,a);
            add(a,b);
            scanf("%d%d",&a,&b);
        }
        for(int i=1;i<=n;i++)
        {
            if(tag[i]==0)
            {
              root=i;
              break;
            }
        }
        dfs(root);
        printf("%d\n",max(dp[root][0],dp[root][1]));
    }
    return 0;
}