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最新推荐文章于 2019-07-15 14:59:24 发布

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探讨了在特定限制条件下，如何通过动态规划算法计算出电影票售卖的最短总时间，以帮助售票员尽早结束工作。

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

2 2 20 25 40 1 8

Sample Output

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

题目大意：一个人上班在售票机那里卖电影票，他可以一次取出一张电影票来卖给别人，也可以一次取出两张来卖给俩人，问这个人最早几点下班（即所有买票人都买完票了）。给了你k个人取票所用时间，k-1组两两组合后取票所用时间。

解题思路：动态规划，状态转移方程为dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i])，即要么第i个人单独买，要么第i个人跟第i-1个人一起买。（上班时间是早上八点）

代码如下：


#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
int a[2010];
int b[2010];
int dp[2010];
int main()
{
    int t;
    int x,y,c;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(dp,0,sizeof(dp));
        int n;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=2; i<=n; i++)
        {
            scanf("%d",&b[i]);
        }
        dp[1]=a[1];
        dp[2]=min(a[1]+a[2],b[2]);//前键
        for(int i=3; i<=n; i++)
        {
            dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);
        }
        int ans=dp[n];
        x=8,y=c=0;                              //时间转换
        y+=(ans/60);
        c=ans%60;
        x+=(y/60);
        y%=60;
        if(x>12||(x==12&&(y>0||c>0)))
            printf("%.2d:%.2d:%.2d pm\n",x,y,c);
        else
            printf("%.2d:%.2d:%.2d am\n",x,y,c);
    }
    return 0;
}