POJ2342 Anniversary party（树状DP）-优快云博客

本文链接：https://blog.youkuaiyun.com/sdfgdbvc/article/details/51510213

本文介绍了一个关于大学周年庆典派对邀请嘉宾的算法题，通过建立员工层级结构并考虑互斥条件，使用动态规划求解最大欢乐值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6190		Accepted: 3567

Description

There is going to be a party to celebrate the 80-th Anniversary of the Ural State University. The University has ahierarchical(分层的)structure(结构) ofemployees(雇员). It means that thesupervisor(监督人) relation forms a tree rooted at therector(校长) 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(人事部门) office hasevaluated(评价)conviviality(欢乐) of each employee, so everyone has some number (rating)attached(依附) to him or her. Your task is to make a list of guests with themaximal(最高的) possible sum of guests' conviviality ratings.

Input

Employees are numbered from 1 to N. A first line ofinput(投入) contains a number N. 1 <= N <= 6 000. Each of thesubsequent(后来的) N lines contains the conviviality rating of the corresponding employee. Conviviality rating is aninteger(整数) 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 h(规格)as the form:
L K
It means that the K-th employee(雇员) is an immediatesupervisor(监督人) of the L-th employee.Input(投入) is ended with the line
0 0

Output

Output(输出) should contain themaximal(最高的) sum of guests' ratings.

Sample Input

Sample Output

题意：有n个人，接下来n行是n个人的价值，再接下来n行给出l，k说的是l的上司是k，这里注意l与k是不能同时出现的

思路：用dp数据来记录价值，开数组用下标记录去或者不去

则状态转移方程为：
DP[i][1] += DP[j][0];
DP[i][0] += max{DP[j][0],DP[j][1]};
其中j为i的孩子节点。这样，从根节点root进行dfs，最后结果为max{DP[root][0],DP[root][1]}。

PS：题目上没说，但是给的关系是从叶子到根的，所以才用root = k;找根。如果不是的话，设置额外数组对每个 l 进行标记，最后找没有被标记的就是根节点。

#include <stdio.h>
#include <string.h>
#define max(a,b) (a)>(b)?(a):(b)
int dp[6005][2], father[6005], vis[6005], N;
void dfs(int x)
{
	vis[x] = 1;
	for(int i = 1; i <= N; i++)
	{
		if(!vis[i] && father[i] == x)
		{
			dfs(i);
			dp[x][1] += dp[i][0];
			dp[x][0] += max(dp[i][0], dp[i][1]);
		}
	}
}
int main()
{
	int i, l, k, root;
	while(scanf("%d", &N) != EOF)
	{
		memset(dp, 0, sizeof(dp));
		memset(vis, 0, sizeof(vis));
		for(i = 1; i <= N; i++)
			scanf("%d", &dp[i][1]);
		root = 0;
		while(scanf("%d%d", &l, &k) && l + k)
		{
			father[l] = k;
			root = k;
		}
		dfs(root);
		printf("%d\n", max(dp[root][0], dp[root][1]));
	}
	return 0;
}