HDU - 1260 Tickets （dp）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1260点击打开链接

Tickets

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5002    Accepted Submission(s): 2614

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
  

 

先讲数据量 因为在50000s以内 因此不会到下午（还是很贴近生活的 一开始没注意都算到可能一直到第二天）

再来是dp过程

用了两个dp 一个为dpone 即最后一个是以单个买票结束 另一个dptwo即最后一个以双人票结束

每次每个状态都是前一个为单票或者双票而推导 

需要注意dptwo的第一个（即一个人）也要做成单人票

#include <bits/stdc++.h>
using namespace std;
void endtime(int s)
{
	int second=s%60;
	int min=s/60;
	int hour=8+(min/60);
	min%=60;
	int flag=1;
	if(hour<10)
		cout << "0";
	cout << hour << ":";
	if(min<10)
		cout << "0";
	cout << min << ":";
	if(second<10)
		cout << "0";
	cout << second << " ";
	cout << "am" << endl;
}
int main()
{
	int t ;
	cin >> t;
	while(t--)
	{
		int n;
		cin >> n;
		if(n==1)
		{
			int mid;
			cin >> mid;
			endtime(mid);
		}
		else 
		{
			int one[2222];
			int two[2222];
			for(int i=1;i<=n;i++)
				cin >> one[i];
			for(int i=2;i<=n;i++)
				cin >> two[i];
			int dpone[2222];
			int dptwo[2222];
			for(int i=1;i<2222;i++)
				dpone[i]=dptwo[i]=INT_MAX;
			dpone[1]=one[1];
			dpone[2]=one[1]+one[2];
			dptwo[1]=one[1];
			dptwo[2]=two[2];
			for(int i=3;i<=n;i++)
			{
				dpone[i]=min(dpone[i-1]+one[i],dptwo[i-1]+one[i]);
				dptwo[i]=min(dpone[i-2]+two[i],dptwo[i-2]+two[i]);
			}
			int ans=min(dpone[n],dptwo[n]);
			endtime(ans);
		}
	}
}