hdu1520Anniversary party（比poj 2342数据强一些）

最新推荐文章于 2022-09-03 14:04:18 发布

原创最新推荐文章于 2022-09-03 14:04:18 发布 · 574 阅读

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本文深入探讨了AI音视频处理领域中的关键技术，特别是视频分割与语义识别。通过详细解释这些技术的工作原理、应用案例及实际效果，旨在为读者提供全面的理解和洞察。此外，文章还涉及了这些技术如何应用于自动驾驶、AR增强现实等场景，展现了AI在音视频领域的广阔应用前景。

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

连接表存储孩子

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

#define N 6005

int dp[N][2],father[N];
int n;
int head[N] ;
int num;

struct stud{
int to;
int next;
}e[N*2];

void add(int s,int to)
{
	e[num].to=to;
	e[num].next=head[s];
	head[s]=num++;
}

void dfs(int x)
{
    int i,j;
    for(i=head[x];i!=-1;i=e[i].next)
	{
		int to=e[i].to;
		dfs(to);

		dp[x][1]+=dp[to][0];
		dp[x][0]+=max(dp[to][1],dp[to][0]);
	}
}

int main()
{
	int i,j;
	while(~scanf("%d",&n))
	{
		memset(dp,0,sizeof(dp));
		
		memset(father,0,sizeof(father));
		memset(head,-1,sizeof(head));

		for(i=1;i<=n;i++)
			scanf("%d",&dp[i][1]);

		int s,e;
		num=0;

		while(scanf("%d%d",&s,&e),s+e)
		{
			father[s]=e;
			add(e,s);
		}

		i=1;
		while(father[i])
			i=father[i] ;

		dfs(i);
		printf("%d\n",max(dp[i][1],dp[i][0]));
	}
	return 0;
}