HDOJ 题目1260 Tickets（动态规划）

最新推荐文章于 2019-02-19 11:31:00 发布

Jogging_Clown

最新推荐文章于 2019-02-19 11:31:00 发布

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分类专栏：动态规划

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本文探讨了在电影院售票场景中，如何通过合理安排相邻顾客同时购买两张票的方式，以最短时间完成所有顾客的购票过程。通过动态规划算法解决实际问题，详细解释了状态转移方程，并给出了具体实现代码及输出格式。

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Tickets

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1193 Accepted Submission(s): 561

Problem Description

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:
1) An integer K(1<=K<=2000) representing the total number of people;
2) K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;
3) (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

Sample Output

  
   08:00:40 am
08:00:08 am

Source

浙江工业大学第四届大学生程序设计竞赛

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题意：有k个人要买票（你也在里面最后一位），每个人买票要花时间，或者两个相邻的人一起买会有一个时间，问你最少花费多少时间才能买到票

思路：题意看的半懂的，结合sample差不多理解了，

以某个人节点，他要么单独买，要么和前面一个人或者后面一个人一起买，

和后面一个人一起买其实可以看成是后面一个人和他前面的一个人一起买，

因此相当于只有两种情况，要么自己一个人买，要么和前面一个人一起买

于是设计dp[i][0],dp[i][1]

0：代表自己一个人买时到他这个位置所需要的最少时间

1：代表自己和前面一个人一起买到他当前位置所需的最少时间

于是状态转移方程就出来了

dp[i][0] = max(dp[i-1][0], dp[i-1][1]) + t[i]; (i >= 1 && i <= n, t表示单独买所需的时间)

dp[i][1] = max(dp[i-2][0], dp[i-2][1]) + interval[i]; (i >= 2 && i <=n, interval表示和前面一个人一起买所需的时间)

最后的answer就是 max(dp[n][0], dp[n][1])了

这个结果是s，换算一下变成时分秒然后加上早上8点整大于12点就对12取模，下午是pm,上午是am

ac代码

#include<stdio.h>
#include<string.h>
int dp[2020][2],t1[2020],t2[2020];
int min(int a,int b)
{
	if(a>b)
		return b;
	else
		return a;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,i,ans,h,m,s;
		scanf("%d",&n);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&t1[i]);
		}
		for(i=2;i<=n;i++)
		{
			scanf("%d",&t2[i]);
		}
		memset(dp,0,sizeof(dp));
		t2[1]=t1[1];
		for(i=1;i<=n;i++)
		{
			dp[i][0]=min(dp[i-1][0],dp[i-1][1])+t1[i];
			if(i==1)
				dp[i][1]=t1[i];
			else
				dp[i][1]=min(dp[i-2][0],dp[i-2][1])+t2[i];
		}
		ans=min(dp[n][0],dp[n][1]);
		h=8+ans/3600;
		ans%=3600;
		m=ans/60;
		ans%=60;
		s=ans;
		if(h>12)
		{
			printf("%02d:%02d:%02d pm\n",h,m,s);
		}
		else
			printf("%02d:%02d:%02d am\n",h,m,s);
	}
}