Tickets

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7318 Accepted Submission(s): 3719

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

题目意思：

    题意：有n个人买票，可以一人买一张，花费时间是a[i]，
    也可以相邻两个人一起买一张两人票，花费时间是b[i]，
    现在求n个人买票需要的最少时间是多少。

分析：

dp方程：

dp[ i ] = min { dp[ i-1 ] + a[ i ] , dp[ i-2 ] + b[ i ] };

对当前人的来说两种选择：就是买一张：dp[ i-1 ] + a[ i ]

或者当前人和前面人一起买（买两张：dp[ i-2 ] + b[ i ]

然后从二种选择里面选出最小的

code：

#include<bits/stdc++.h>
using namespace std;
#define max_v 2005
int a[max_v];
int b[max_v];
int dp[max_v];
int main()
{
    /*
    题意：有n个人买票，可以一人买一张，花费时间是a[i]，
    也可以相邻两个人一起买一张两人票，花费时间是b[i]，
    现在求n个人买票需要的最少时间是多少。
    分析：
    dp[ i ] = min { dp[ i-1 ] + a[ i ] , dp[ i-2 ] + b[ i ] };
    */
    int t,n;
    int h,f,s;
    cin>>t;
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        for(int i=2;i<=n;i++)
        {
            cin>>b[i];
        }
        dp[0]=0;
        dp[1]=a[1];
        for(int i=2;i<=n;i++)
        {
            dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);
        }
        s=dp[n]%60;
        f=dp[n]/60%60;
        h=dp[n]/60/60;
        h+=8;
        h%=24;
        if(h>12)
        {
            h-=12;
            printf("%02d:%02d:%02d pm\n",h,f,s);
        }else
        {
            printf("%02d:%02d:%02d am\n",h,f,s);
        }

    }
    return 0;
}

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