杭电 1260 Tickets

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

大意：
　　 卖电影票，先输入t，代表t组测试数据，没组测试数据，第一行输入一个n代表n个人，第二行为n个整数代表票卖给每个人用的时间，第二行为n-1个数，代表卖给两个人（第i个人和第i+1个人，与上面的人对应）　　 用的时间；
　　 例如：
　　　　　　3
　　　　　　a[1] a[2] a[3]
　　　　　　b[1] b[2]
　　　a[i]代表卖票给第i个人用的时间，b[i]代表同时卖票给第i个人和第i+1个人用的时间。
思路：
　　定义数组a[i]记录单个人买票的时间b[i]记录两个人买票的时间，d[i]记录i个人买票的最短时间，根据
　　d[i]=min(d[i-2]+b[i-1],d[i-1]+a[i])（i-2个人用的最短时间加第i-1和第i个人一起买票与i-1个人用的最短时间加第i个人买票的时间），   求出n个人买票用的最短时间，最后控制输出格式。

#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int a[2010],b[2010],d[2020];
        int n;
        scanf("%d",&n);
        for(int i = 1 ; i <= n ; i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i = 1 ; i <= n-1 ; i++)
        {
            scanf("%d",&b[i]);
        }
        d[0]=0;
        d[1]=a[1];
        for(int i = 2 ; i <= n ; i++)
        {
            d[i]=min(d[i-2]+b[i-1],d[i-1]+a[i]);        //表示i个人买票用的最短时间 
        }
        int s=d[n]%60;
        int m=d[n]/60%60;
        int t=8+d[n]/60/60;
        printf("%02d:%02d:%02d am
",t,m,s);
    }
}