HDU1260Tickets（dp）

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：一个人卖票，他可以同时卖给两个人票，也可以只卖给一个人票。给你他卖给一个人票的时间和卖给相连两个人票的时间，他八点上班，问什么时候下班。

思路：用dp做。判断min(dp[i-1]+a[i],dp[i-2]+b[i-1]) 就好了 a[i]是单独卖花费时间 b[i]是两个一起卖花费时间。这两个一个是 单独买第i个，一个是两个一起买。因为不管dp[i-2]是单独买的还是和上一个一起买的都不影响下两个一起买，dp[i-1]也不会影响下一个单独买。知道最后一个。输出比较麻烦，首先根据秒数算出小时 分钟 。在判断时间是否超过了十二点，超过了最后是pm。其次就是输出0的问题了，分开讨论，如果该时刻（时针，分针，秒针）大于时直接输出这个时间，如果小于线输出0在输出时间。就是这个道理了。写的时候我在想给的数据会不会到第二天，就是时间会不会超过24小时。但根据常理来说一天上下班不可能到第二天，所以就不用考虑到第二天的情况了！

代码：

#include  
using namespace std;  
int a[10010]; 
int b[10010]; 
int dp[10010];  
int main()  
{  
    int t,k=1;  
    scanf("%d",&t);
    while(t--)
    {  
    	memset(dp,0,sizeof(dp));
    	memset(a,0,sizeof(a));
    	memset(b,0,sizeof(b));
    	int n,sum=0;
    	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]);
		dp[1]=a[1];
		for(int i=2;i<=n;i++)
		{
		dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i-1]);
		}
		int h,f,m;
		h=dp[n]/3600;
		f=(dp[n]-3600*h)/60;
		m=(dp[n]-3600*h-60*f);
		h+=8;
		if(h>12)
		{
			printf("%d:",h);
			if(f>=10)
			printf("%d:",f);	
			else
			printf("0%d:",f);
			if(m>=10)
			printf("%d pm\n",m);
			else
			printf("0%d pm\n",m);
		}
		else
		{
			if(h>=10)
			printf("%d:",h);
			else
			printf("0%d:",h);
			if(f>=10)
			printf("%d:",f);	
			else
			printf("0%d:",f);
			if(m>=10)
			printf("%d am\n",m);
			else
			printf("0%d am\n",m);
		}
    }  
    return 0;  
}