zoukankan      html  css  js  c++  java
  • [dp专题]hdu 1260 tickets

    Tickets

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 1408    Accepted Submission(s): 687


    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
     
    Recommend
    JGShining   |   We have carefully selected several similar problems for you:  1257 1231 1074 1069 1159 
     
     
    dp
    状态转移方程为dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i])   dp[i]为第i个人买完票所花的最少时间。a[i]和b[i]分别为单独买票和两个人一起买票的时间。
    因为调试语句没删掉和错把B[I]的下表认为是从1开始(实际应从2开始) WA了两发 3A
     
    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cmath>
    #include <cstring>
    using namespace std;
    
    int n,k;
    const  int inf=99999999;
    const int N=2E3+5;
    bool am;
    int a[N],b[N],dp[N],hour,mini,sec;
    
    int main()
    {
    
        scanf("%d",&n);
        while ( n-- )
        {
            am = true;
            memset(a,0,sizeof(a));
            memset(b,0,sizeof(b));
            hour = 8;
            mini = 0;
          //  cout<<"hour1:"<<hour<<endl;
            for ( int i = 0 ; i < N ; i++ )
                dp[i] = 0;
         //   cout<<"hour2:"<<hour<<endl;
    
            scanf("%d",&k);
            for ( int i = 1; i <= k ; i++ )
                scanf("%d",&a[i]);
            for ( int i = 2 ; i <= k ; i++ )
                scanf("%d",&b[i]);
            dp[1] = a[1];
            for ( int i = 2; i <= k ; i++ )
                dp[i] = min(dp[i-1]+a[i],dp[i-2]+b[i]);
        //    printf("%d
    ",dp[k]);
            sec = dp[k];
         //   hour = 8;
       //  cout<<"hour:"<<hour<<endl;
            while (sec>=3600)
            {
                sec = sec - 3600 ;
                hour++;
            }
            while (sec>=60)
            {
                sec = sec - 60 ;
                mini++;
            }
            if ( hour>12 )
            {
                am = false;
                hour = hour - 12;
            }
            if ( hour<10 )
                cout<<"0";
            cout<<hour<<":";
            if ( mini<10 )
                cout<<"0";
            cout<<mini<<":";
            if ( sec<10 )
                cout<<"0";
            if ( am )
                cout<<sec<<" am"<<endl;
            else cout<<sec<<" pm"<<endl;
        }
        return 0;
    }
     
  • 相关阅读:
    Android Lock Screen Orientation
    android studio修改新项目package名称
    DevExpress控件GridView挂下拉控件无法对上值
    Datawindow.net+access数据窗口制作方法
    LR11中web_save_timestamp_param
    LR自带网站飞机订票系统 启动
    MySQL ODBC驱动安装和配置数据源
    【笔记】LR配置ODBC连接数据库进行参数化(mysql )未完待续
    【笔记】LR录制方式和常用函数
    【笔记】http1.1支持的7种请求方法
  • 原文地址:https://www.cnblogs.com/111qqz/p/4397233.html
Copyright © 2011-2022 走看看