卖电影票求最短时间问题（动态规划）

最新推荐文章于 2022-11-13 09:24:35 发布

原创最新推荐文章于 2022-11-13 09:24:35 发布 · 470 阅读

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#动态规划

该博客探讨了一个关于售票员如何通过动态规划方法优化售票过程的问题。内容涉及到一个名为Jesus的售票员，他的任务是在最短时间内售完电影票。每个顾客购票所需时间和相邻顾客一起购票所需时间被给出。通过动态规划算法，可以确定最短的售票时间，从而让Jesus尽早下班。示例输入和输出展示了如何计算这个最短时间，并将其转换为易于理解的时间格式。

Jesus, what a great movie! Thousands of people are rushing to the cinema. However, this is really a tuff time for Joe who sells the film tickets. He is wandering when could he go back home as early as possible.
A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together. As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time.
Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her. Could you so kind to tell poor Joe at what time could he go back home as early as possible? If so, I guess Joe would full of appreciation for your help.
Input
There are N(1<=N<=10) different scenarios, each scenario consists of 3 lines:

An integer K(1<=K<=2000) representing the total number of people;
K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;
(K-1) integer numbers(0s<=Di<=50s) representing the time needed for two adjacent people to buy two tickets together.
Output
For every scenario, please tell Joe at what time could he go back home as early as possible. Every day Joe started his work at 08:00:00 am. The format of time is HH:MM:SS am|pm.
Sample Input
2
2
20 25
40
1
8
Sample Output
08:00:40 am
08:00:08 am

题目大意就是说：Jesus是个售票员早上8点上班卖完票就可以走人，顾客可以单独买票也可以和相邻人一起买，如果有k个顾客，怎么能最快卖完回家，求最快回家时间。

详情看注释

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
	int n,k,i,j,h,m,s;
	int a[2010],b[2010];//作用略，看题目
	int dp[2010];
	scanf("%d",&n);
	while(n--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		scanf("%d",&k);
		for(i=1;i<=k;i++)
			scanf("%d",&a[i]);
			
		for(i=2;i<=k;i++)
			scanf("%d",&b[i]);
			
		dp[1]=a[1];	
		for(i=2;i<=k;i++)
		{
			//走过i，i之前的已经是短时间 
			dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);
			/*理解： 
			每次考虑两种情况：1.让i单独买----dp[i-1](i-1之前的最短时间)加a[i]（i单独买票的时间）  
			   					2.让i和前一个人一起买----此时要解放第i-1个人，因此是dp[i-2]（i-2之前的最短时间 ）加b[i](i与i-1一起买票的时间) 
			*/
		}
		s=dp[k]%60;
		m=dp[k]/60%60;
		h=dp[k]/3600;
		h+=8;
		if(h>12)
			printf("%02d:%02d:%02d pm\n",h,m,s);
		else
			printf("%02d:%02d:%02d am\n",h,m,s);
		
	}
	return 0;
}

动态规划dp