hdu1520 Anniversary party-优快云博客

本文介绍了一种使用树形动态规划解决员工聚会问题的方法，旨在最大化参与者的总评分，同时遵循特定的企业层级结构限制。

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7191 Accepted Submission(s): 3174

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

Sample Output

5

这题是一道树形dp，可以用dp[i][0]表示以这个节点为根的并且这个节点取的所得到的最大价值，用dp[i][1]表示以这个节点为根的并且这个节点不取取的所得到的最大价值，然后用dfs搜一下就行了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define maxn 6006
int w[maxn],dp[maxn][2],pre[maxn],flag[maxn];
int first[maxn];
struct node{
    int to,next;
}e[2*maxn];

void dfs(int u)
{
    int i,j;
    if(first[u]==-1){
        dp[u][0]=w[u];
        dp[u][1]=0;return;
    }
    dp[u][0]=w[u];dp[u][1]=0;
    for(i=first[u];i!=-1;i=e[i].next){
        int v=e[i].to;
        dfs(v);
        dp[u][1]=max(dp[u][1],dp[u][1]+max(dp[v][0],dp[v][1]) );
        dp[u][0]=max(dp[u][0],dp[u][0]+dp[v][1]);
    }
}

int main()
{
    int n,m,i,j,c,d;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++){
            scanf("%d",&w[i]);
            pre[i]=i;
        }
        memset(first,-1,sizeof(first));
        memset(dp,0,sizeof(dp));
        int tot=0;
        while(scanf("%d%d",&c,&d)&&(c||d)){
            tot++;
            e[tot].to=c;e[tot].next=first[d];
            first[d]=tot;
            pre[c]=d;
        }
        int ans=0;
        for(i=1;i<=n;i++){
            if(pre[i]==i){
                dfs(i);
                ans+=max(dp[i][0],dp[i][1]);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}