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


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

2
2
20 25
40
1
8

 

[align=left]Sample Output[/align]

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;

 

 }

 }

}