Anniversary party POJ - 2342

最新推荐文章于 2021-08-04 15:15:41 发布

转载最新推荐文章于 2021-08-04 15:15:41 发布 · 99 阅读

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原文链接：http://www.cnblogs.com/xfww/p/8444532.html

本文介绍了一种使用树形动态规划(DP)的方法来解决一个特定的问题：即在一个有层级结构的大学中，如何选择参与80周年校庆派对的员工名单，使得这些员工的“亲和力”评分总和最大，同时确保没有员工与其直接上司同时参加。

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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

Sample Output

5

中文题目:
一棵树有n个结点，每个结点有一个权值，一个结点和其直接的父亲结点不能同时选，求按照要求所能选到的点权和的最大值

做法：
简单的DP

#include<iostream>
using namespace std;
#include<cstdio>
#include<cstring>
#include<vector>
typedef long long ll;
ll dp[10010][2];
ll w[10010];
vector<int> maps[10010];
void init(int n){//初始化
	for(int i=0;i<=n;i++){
		 maps[i].clear();
		 dp[i][0]=dp[i][1]=-1;
		 w[i]=0;
	}
}
ll dfs(int x,int y){
//	cout<<"***L: "<<x<<" "<<y<<endl;
	if(x<=0)
		return 0;
	if(dp[x][y]!=-1)
		return dp[x][y];
	ll ans = 0;
	for(int i=0;i<maps[x].size();i++){
		int p = maps[x][i];
	//	cout<<p<<endl;
		if(y==0)
			ans +=dfs(p,1);
		else{
			ll pp = dfs(p,0);
			ll qq = dfs(p,1);
			ans +=max(pp,qq);
		}
	}
	if(y==0)
		ans+=w[x];
	return dp[x][y]=ans;
}
int main(){
	int n;
	while(scanf("%d",&n)!=EOF){
		if(n==0)
			break;
		init(n);
		for(int i=1;i<=n;i++)
			scanf("%lld",&w[i]);
		for(int i=1;i<n;i++){
			int p,q;
			scanf("%d%d",&p,&q);
			maps[q].push_back(p);
		}
		ll ans = 0;
		for(int i=1;i<=n;i++){
			ans = max(ans,dfs(i,0));
			ans = max(ans,dfs(i,1));
		}
		printf("%lld\n",ans);
	}
	return 0;
}

转载于:https://www.cnblogs.com/xfww/p/8444532.html