HDU1260 Tickets

Tickets

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1824 Accepted Submission(s): 883


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


Source

浙江工业大学第四届大学生程序设计竞赛

题意：给定N个数，1-N依次代表每张票卖出所需时间。在给定N-1个数，1-(N-1)依次代表每相邻两张票卖出所需时间。

求最少卖完时间。时间从8点整开始，注意am/pm。

思路：数组a[n] , b[n-1] 。

动态转移方程 dp[i] = min(dp[i-1]+a[i],dp[i-2]+b[i-1])  dp[i]意为对于第i张票所化最少时间。

有两种决策：1 单卖第i张 ；2 双卖第i和第i-1张（b数组下标对应为i-2,i-1)

#include <iostream>#include <cstdio>#include <cstring>#define max(a,b) (a>b?a:b)#define min(a,b) (a>b?b:a)const int maxn = 2000+100 ;const int inf = 1<<30 ;using namespace std;int main() {    int ncase, n  ;    int a[maxn] , b[maxn] ;    int dp[maxn] ;    cin >> ncase ;    while(ncase--) {        scanf("%d",&n) ;        for( int i = 1 ; i <= n ; ++i ) {            scanf("%d",&a[i]) ;        }        for( int j = 1; j <= n-1 ; ++j ) {            scanf("%d",&b[j]) ;        }        memset(dp,0,sizeof(dp)) ;        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 second = dp[n]%60 ;        int minutes = (dp[n]/60)%60 ;        int hour = dp[n]/3600+8 ;        if(dp[n] <= 3600*4)        printf("%02d:%02d:%02d am\n",hour,minutes,second) ;    }    return 0 ;}