hdoj 1260 Tickets【dp】

Tickets

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1935    Accepted Submission(s): 933

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

 

很久没做dp了  再加上自己dp本来就很渣，下午比赛时看人家一个一个都做出来，自己只能眼巴巴的看着，唉！！！智商啊！！

题意：一群人去买票，先输入每个人单独买票所花费的时间，在给出两个人两两结合买票所花费的时间，求最短时间

题解：需要推出状态转移方程，设数组a[]是单个人买票所花费的时间，数组b[]是两个人一起买票所花费的时间，dp[i]表示

        前i个人买票所花费的时间，则状态转移方程是：dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);

#include<stdio.h>
#include<string.h>
#define MAX 2100
#define min(x,y)(x<y?x:y)
int a[MAX],b[MAX],dp[MAX];
int main()
{
	int t,i,j,n;
	int h,m,s;
	int sum,tot;
	scanf("%d",&t);
	while(t--)
	{
		memset(dp,0,sizeof(dp));
		scanf("%d",&n);
		for(i=1;i<=n;i++)
		    scanf("%d",&a[i]);
		for(i=2;i<=n;i++)
		    scanf("%d",&b[i]);
		dp[1]=a[1];
	   for(i=2;i<=n;i++)
	   	  dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);
	   	  //printf("%d\n",dp[n]);
	   sum=dp[n];
	   h=0;s=0;m=0;
	   s=sum%60;
	   m=(sum-s)/60;
	   if(m>=60)
	   {
	   	   h=h+m/60;
	   	   m=m%60;
	   }
	   h=8+h;
	   if(h<=12) 
	   printf("%02d:%02d:%02d am\n",h,m,s);
	   else
	   {
	   	 h-=12;
	   	 printf("%02d:%02d:%02d pm\n",h,m,s);
	   }
	   
    }
	return 0;
}