【HDU】-1260-Tickets（简单DP）

Tickets

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3148    Accepted Submission(s): 1545

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

2220 254018

 

Sample Output

08:00:40 am08:00:08 am

 

题意：有n张票，可以一次卖一张，也可以一次卖两张，分别知道卖一张和两张的时间，求最快卖完的时间。

题解：简单DP，理解。

每次卖票都有两种情况，卖一张或者两张，取其中较短的时间即可。

注意输出，开始就是因为输出wa了一下。

#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<stack>#include<algorithm>using namespace std;#define CLR(a,b)  memset(a,b,sizeof(a))int main(){int u;scanf("%d",&u);while(u--){int n;int a[2020],b[2020],dp[2020];CLR(a,0),CLR(b,0),CLR(dp,0);scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);for(int i=1;i<n;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,m,s;h=dp[n]/3600;m=(dp[n]%3600)/60;s=(dp[n]%3600)%60;h+=8;if(h>=12)printf("%02d:%02d:%02d am\n",h,m,s);elseprintf("%02d:%02d:%02d am\n",h,m,s);}return 0;}