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最新推荐文章于 2020-02-07 11:05:28 发布

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本文介绍了一个售票场景下的优化问题，通过动态规划算法找到使售票员能够尽早下班的最佳方案。顾客可以选择单独购票或两人一同购票，算法需计算每种情况下所需的时间。

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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

啊，想象一下，复仇者联盟4上映的时候影院门口排起了长龙，有的顾客很迅速一会就买好票了，有的就很墨迹要半天。二你作为一个售票员，当然要想一种可以让你自己最快下班的方法啦。顾客可以一个人买票，也可以两个人一起买票。当然每个时间都不一样啦。

用one数组统计每个人自己买要多少时间，用two数组统计和前面的人一起买要多少时间。dp数组表示到这个人为止的最小时间。这样最优解就是对应的dp数组加上one[i]或者two[i]的最小值，由于只要考虑前面两个人，所以只要一个循环就OK啦。

总觉得最长时间会超过四小时到下午，但实际上我没写判断am pm也过了，大概是数据不严格的原因吧。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int one[2010],two[2010],dp[2010];
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        int k;
        cin>>k;
        memset(one,0,sizeof(one));
        memset(two,0,sizeof(two));
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=k;i++)
            scanf("%d",&one[i]);
        for(int i=2;i<=k;i++)
            scanf("%d",&two[i]);
        dp[0]=0;
        dp[1]=one[1];
        for(int i=2;i<=k;i++)
            dp[i]=min(dp[i-1]+one[i],dp[i-2]+two[i]);
        int h=dp[k]/3600+8;
        int m=dp[k]/60%60;
        int s=dp[k]%60;
        printf("%02d:%02d:%02d am\n",h,m,s);

    }
    return 0;
}