hdu 1260 Tickets

Tickets

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 751    Accepted Submission(s): 379

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

 

这道题是说 有k个人买票。如果相邻的两个一起买票 就可以少花一些时间。求最短的卖票时间。

用s数组存储一个买票所需要花的时间。

用ss数组存储两个买票所花的时间。

那么状态转移方程就是opt[i]=min(opt[i-1]+s[i],opt[i-2]+ss[i-1]);

也就是说   第一种情况：.  第i人单独买+前i-1买票所需时间   第二种情况是 第i个人和第i-1人一起买+前i-2人买票所需要的最少时间。

 

只有多做 多刷 自己才能更明白更理解。

 

#include<iostream>
#include<iomanip>
using namespace std;
int ss[3000];
int s[3000];
int opt[3000];
int main()
{
 int cas,k,i,j,l,t,h,m,second;
 cin>>cas;
 while(cas--)
 {
 s[0]=0;
 cin>>k;
 for(i=1;i<=k;i++)
  cin>>s[i];
 ss[0]=0;
 for(j=1;j<=k-1;j++)
  cin>>ss[j];

 opt[0]=0;
 opt[1]=s[1];
 for(l=2;l<=k;l++)
 {
 opt[l]=opt[l-1]+s[l]<opt[l-2]+ss[l-1]?opt[l-1]+s[l]:opt[l-2]+ss[l-1];
 
 }
 
 t=opt[k];

 second=t%60;
 t=(t-second)/60;
 m=t%60;
 t=(t-m)/60;
 h=8+t;

 if(h<12)
 {
 cout<<setfill('0')<<setw(2)<<h<<':'<<setfill('0')<<setw(2)<<m<<':'<<setfill('0')<<setw(2)<<second<<" am"<<endl;

 }
 else
 {
 cout<<setfill('0')<<setw(2)<<h-12<<':'<<setfill('0')<<setw(2)<<m<<':'<<setfill('0')<<setw(2)<<second<<" pm"<<endl;
 
 }
 }

}