PAT A1016.Phone Bills

A long-distance telephone company charges its customers by the following rules:

Making a long-distance call costs a certain amount per minute, depending on the time of day when the call is made. When a customer starts connecting a long-distance call, the time will be recorded, and so will be the time when the customer hangs up the phone. Every calendar month, a bill is sent to the customer for each minute called (at a rate determined by the time of day). Your job is to prepare the bills for each month, given a set of phone call records.

Input Specification:

Each input file contains one test case. Each case has two parts: the rate structure, and the phone call records.

The rate structure consists of a line with 24 non-negative integers denoting the toll (cents/minute) from 00:00 - 01:00, the toll from 01:00 - 02:00, and so on for each hour in the day.

The next line contains a positive number N (<= 1000), followed by N lines of records. Each phone call record consists of the name of the customer (string of up to 20 characters without space), the time and date (mm:dd:hh:mm), and the word "on-line" or "off-line".

For each test case, all dates will be within a single month. Each "on-line" record is paired with the chronologically next record for the same customer provided it is an "off-line" record. Any "on-line" records that are not paired with an "off-line" record are ignored, as are "off-line" records not paired with an "on-line" record. It is guaranteed that at least one call is well paired in the input. You may assume that no two records for the same customer have the same time. Times are recorded using a 24-hour clock.

Output Specification:

For each test case, you must print a phone bill for each customer.

Bills must be printed in alphabetical order of customers' names. For each customer, first print in a line the name of the customer and the month of the bill in the format shown by the sample. Then for each time period of a call, print in one line the beginning and ending time and date (dd:hh:mm), the lasting time (in minute) and the charge of the call. The calls must be listed in chronological order. Finally, print the total charge for the month in the format shown by the sample.

Sample Input:

10 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 1010CYLL 01:01:06:01 on-lineCYLL 01:28:16:05 off-lineCYJJ 01:01:07:00 off-lineCYLL 01:01:08:03 off-lineCYJJ 01:01:05:59 on-lineaaa 01:01:01:03 on-lineaaa 01:02:00:01 on-lineCYLL 01:28:15:41 on-lineaaa 01:05:02:24 on-lineaaa 01:04:23:59 off-line

Sample Output:

CYJJ 0101:05:59 01:07:00 61 $12.10Total amount: $12.10CYLL 0101:06:01 01:08:03 122 $24.4028:15:41 28:16:05 24 $3.85Total amount: $28.25aaa 0102:00:01 04:23:59 4318 $638.80Total amount: $638.80

/*************************侯尼玛*************************/

一道简单，但是看起来感觉乱乱的题目。

我的做法是把每一条信息作为一个结构体放在一起，然后用algorithm头文件附带的sort函数排序。sort真是非常好用的函数，可以在参数里引用函数来自定义排序方式。不过这做法似乎是面向过程编程的遗毒之一，其实不符合c++的精神？

因为题目要求是找最靠近的配对记录，因此在排序完成之后，只要找其中连续且配对的记录，就可以输出了。

不过一个要注意的点是：如果一个顾客没有可以配对的两条记录的话，他是不应该出现在输出里的。这个错误一开始困扰了我很久。

附上代码：

    #include <iostream>    #include <stdio.h>    #include <string>    #include <vector>    #include <algorithm>    using namespace std;    double Toll[24]={0};    typedef struct record    {        string customer;        int day;        int hour;        int minute;        bool start;    }Record;    bool Compare(const Record &A,const Record &B)    {        if(A.customer!=B.customer)            return A.customer<B.customer;        else if(A.day!=B.day)            return A.day<B.day;        else if(A.hour!=B.hour)            return A.hour<B.hour;        else            return A.minute<B.minute;    }    double Cost(const Record A,const Record B,int &time)    {        double money=0;        time = (B.day-A.day)*24*60+(B.hour-A.hour)*60+B.minute-A.minute;        /**money+=(60-A.minute)*Toll[A.hour];        for(int j=A.hour+1;j<=B.hour;j++)            money+=60*Toll[j];        money-=(60-B.minute)*Toll[B.hour];        if(B.day>A.day)            for(int i=0;i<B.day-A.day;i++)                for(int k=0;k<24;k++)                    money+=60*Toll[k];**/        int k=0;        for(k=0;k<24;k++)            money+=60*Toll[k]*(B.day-A.day);        for(k=0;k<B.hour;k++)            money+=60*Toll[k];        money+=B.minute*Toll[B.hour];        for(k=0;k<A.hour;k++)            money-=60*Toll[k];        money-=A.minute*Toll[A.hour];        return money;    }    int main()    {        for(int i= 0;i<24;i++)        {            cin>>Toll[i];            Toll[i]/=100;        }        int RecordNum;        cin>>RecordNum;        vector<Record> RCD;        int Month;        for(int i=0;i<RecordNum;i++)        {            Record Temp;            char CharTemp;            cin>>Temp.customer>>Month>>CharTemp>>Temp.day>>CharTemp>>Temp.hour>>CharTemp>>Temp.minute;            string OnOff;            cin>>OnOff;            if(OnOff[1]=='n')                Temp.start=true;            else Temp.start=false;            RCD.push_back(Temp);        }        sort(RCD.begin(),RCD.end(),Compare);        double sum=0;        for(int i=0;i<RecordNum-1;i++)        {            /*if(i==0 || RCD[i].customer!=RCD[i-1].customer)            {                printf("%s %02d\n",RCD[i].customer.c_str(),Month);            }*/            if(RCD[i].start==true && RCD[i+1].start==false && RCD[i].customer==RCD[i+1].customer)            {                if(sum==0) printf("%s %02d\n",RCD[i].customer.c_str(),Month);                printf("%02d:%02d:%02d %02d:%02d:%02d ",RCD[i].day,RCD[i].hour,RCD[i].minute,RCD[i+1].day,RCD[i+1].hour,RCD[i+1].minute);                int time;                double money;                money = Cost(RCD[i],RCD[i+1],time);                sum+=money;                printf("%d $%.2f\n",time,money);            }            if((i==RecordNum-2 || RCD[i].customer!=RCD[i+1].customer) && sum!=0)            {                printf("Total amount: $%.2f\n",sum);                sum=0;            }        }        /***for(int i=0;i<RecordNum;i++)            cout<<RCD[i].customer;***/        return 0;    }