POJ - 2342 D - Anniversary party_there is going to be a party to celebrate the 80-t-优快云博客

博客围绕大学聚会最大欢乐度问题展开，每个人有欢乐度且不想和上司碰面。将上下级关系看作树，通过动态规划求解，设dp[i][0]表示不选i，dp[i][1]表示选i，给出状态转移方程，还提及输入输出要求。

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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
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
题意：一个大学聚会，每个人有一个自己的欢乐度，每个人都不想和自己的上司碰面，问最大的欢乐度是多少。

分析：上下级关系可以看成是一棵树，设dp[i][0]表示不选i这个人，dp[i][1]选i这个人，那么dp[v][1]+=dp[i][0]; dp[v][0]+=max(dp[i][0],dp[i][1]);

代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int book[6010];
int vis[6010];
int f[6010];
int dp[6010][2];
int n;
void dfs(int v)  //dp
{
    book[v]=1;
    for(int i=1; i<=n; i++)
    {
        if(!book[i]&&f[i]==v)  //找到下级
        {
            dfs(i);
            dp[v][1]+=dp[i][0];
            dp[v][0]+=max(dp[i][0],dp[i][1]);
        }
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        int a,b,c;
        memset(book,0,sizeof(book));
        memset(vis,0,sizeof(vis));
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=n; i++)
            scanf("%d",&dp[i][1]);  //记录欢乐度
        while(~scanf("%d%d",&b,&c),b,c)
        {
            f[b]=c;                  //标记上下级
            vis[b]=1;
        }
        for(int i=1; i<=n; i++)
            if(!vis[i])             //找到没有上级的
                a=i;
        dfs(a);
        printf("%d\n",max(dp[a][0],dp[a][1]));
    }
}