怒刷DP之 HDU 1260

最新推荐文章于 2024-05-22 17:32:19 发布

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原文链接：http://www.cnblogs.com/xz816111/p/4789399.html

本文介绍了一种通过优化售票流程来减少顾客等待时间的方法。针对特定的售票机限制，算法探讨了如何让相邻顾客一起购买电影票以缩短总售票时间，并提供了一个C++实现示例。

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Tickets

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1260

Appoint description: System Crawler (2015-09-06)

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

2 2 20 25 40 1 8

Sample Output

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

#include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <climits>
using    namespace    std;

const    int    SIZE = 2005;
int    DP[SIZE],S[SIZE],H[SIZE];

int    main(void)
{
    int    t,n;

    scanf("%d",&t);
    while(t --)
    {
        memset(DP,0,sizeof(DP));
        scanf("%d",&n);
        for(int i = 1;i <= n;i ++)
            scanf("%d",&S[i]);
        for(int i = 1;i <= n - 1;i ++)
            scanf("%d",&H[i]);

        if(n == 1)
            printf("08:00:%02d am\n",S[1]);
        else
        {
            for(int i = 1;i <= n;i ++)
                DP[i] = min(DP[i - 1] + S[i],DP[i - 2] + H[i - 1]);
            int    sec = DP[n];
            int    hour = sec / (60 * 60);
            sec -= hour * 60 * 60;
            int    min = sec / 60;
            sec -= min * 60;
            hour += 8;

            bool    flag_am = true;
            if(hour >= 12)
            {
                flag_am = false;
                hour -= 12;
            }
            printf("%02d:%02d:%02d %s\n",hour,min,sec,flag_am ? "am" : "pm");
        }
    }

    return    0;
}