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.

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.

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.

2
2
20 25
40
1
8

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

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解题思路：

一个人可以单独买票花费一定的时间，也可以两个人一起买票，也给定一个时间，

给出K个人的单独买票时间和K-1个相邻的两个人一起买票的时间，问一共花费的最小时间。

map[]表示一个人单独买的时间，v[]表示两个人一起买的时间。
决策就是，这两个人是一起买还是分开买；

代码：0MS

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 2050
#define INF 999999
int map[M],v[M],dp[M];
int main()
{
    int i,n,m,t,j,k;
    while(cin>>t)
    {
        while(t--)
        {
            cin>>n;
            for(i=1;i<=n;i++)
            cin>>map[i];            //单独买票的时间。
            for(i=1;i<=n-1;i++)
            cin>>v[i];              //一起买票的时间。
            for(i=0;i<=n;i++) dp[i]=(i==0?0:INF);
            dp[1]=map[1];
            for(i=2;i<=n;i++)
            {
                dp[i]=min(dp[i-1]+map[i],dp[i-2]+v[i-1]);  //决策。
            }
        int h = dp[n] / 3600 + 8;         //将时间转换。
        int m = dp[n] / 60 % 60;
        int s = dp[n] % 60;
        printf("%02d:%02d:%02d am\n", h, m, s);  //没有两位的补足前到零。
        }
    }
    return 0;
}