HDU - 1260 Tickets（线性DP）

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

题意：有K个人排队购买电影票，每个人可以选择自己购买或者和相邻的人组成两人小队一起购买，单独购买的时间为si，组队购买从第2个人开始，和前一个人一起购买花费的总时间是di（时间单位为秒）。8点开始售票，问所有人买到票的最早结束时间（12小时制，上午为am，下午为pm）。

1. 状态表示：dp[i]表示前i个人买票花费最少总时间
2. 动态转移：当第i个人买票时，有两种选择，自己单独购买或者和前一个人组队购买
3. 单独购买：dp[i-1] + s[i](s[i]表示第i个人单独购买花费时间)
4. 组队购买：dp[i-2] + d[i](d[i]表示第i个人和前一个人组队购买花费时间)
5. 状态转移方程dp[i] = min(dp[i-1] + s[i], dp[i-2] + d[i])

#include<iostream>
using namespace std;

const int N = 2010;
int dp[N],a[N],b[N];

int main()
{
    
	int T;scanf("%d",&T);
	while(T--)
	{
    
		int k;scanf("%d",&k);
		for(int i=1;i<=k;i++) scanf("%d",&a[i]);
		for(int i=2;i<=k;i++) scanf("%d",&b[i]);
		dp[1]=a[1],dp[2]=min(a[1]+a[2],b[2]);
		for(int i=3;i<=k;i++)
			dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);
		int h=8+dp[k]/3600,m=dp[k]%3600/60,s=dp[k]%60;
		printf("%02d:%02d:%02d am\n",h,m,s);
	}
	return 0;
}