POJ - 2342 （2020.5.13D题）-优快云博客

本文介绍了一个树形动态规划问题，旨在为大学80周年庆典制定一份员工派对名单，使得总欢乐指数最大，同时避免直接上下级同时出席。通过构建员工间的层级关系树，使用深度优先搜索(DFS)算法，实现对每个节点两种状态的最优选择。

Problem
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.

典型的树形dp，不多解释

AC代码

#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;

const int MAXN = 6010;
int dp[MAXN][2];
int visit[MAXN];
int father[MAXN];

int N;
int i, j;
int num, a, b;
int root=0;
void dfs_root(int node)
{
	visit[node] = 1;
	for (int i = 1; i <= N; i++)
	{
	
		if (!visit[i] && father[i] == node)
		{
		
			dfs_root(i);
			dp[node][0] += max(dp[i][1], dp[i][0]);
			dp[node][1] += dp[i][0];
		}
	}
}

int main()
{
	cin >> N;
	for (i = 1; i <= N; i++)
	{
	
		cin >> dp[i][1];
	}
	
	while (cin >> a >> b)
	{
	
		if (a + b == 0)
			break;
		father[a] = b;
	}
	/*for (i = 1; i <= N; i++)
	{

		cout << "*" << father[i] << endl;
	}*/
	root = 1;
	while (father[root])
	{
	
		root = father[root];
	}
	dfs_root(root);
	
	cout << max(dp[root][0], dp[root][1]) << endl;
}