PAT刷题：1016. Phone Bills (25)

本题有个大坑：如果客户A的账单记录没有任何一对是合法的，则客户A什么信息都不输出，对于这一点，出题者不说明，也不容易测试出来，我只想问候他全家。

判读有效记录，只要本条记录是on-line，下一条记录是off-line，就是有效记录

先计算出两头零碎时间，再计算出中间通话多少小时，用循环相乘得到结果

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 $638.80
Total amount: $638.80
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <functional>#include <numeric>#include <vector>#include <list>#include <map>#include <string>#include <queue>using namespace std;struct Record{    string name;    string time;    string month;    string date;    int day;    int hour;    int minute;    string tag;    Record() : name("") , time("") , tag(""){}    Record(string _name , string _time , string _tag) : name(_name) , time(_time) , tag(_tag) {        month=time.substr(0,2);        date=time.substr(3);        day = atoi(time.substr(3,2).c_str());        hour = atoi(time.substr(6,2).c_str()) ;        minute=atoi(time.substr(9,2).c_str());    }    bool operator < (const Record &x)const {        if(name != x.name)return name<x.name;        else if(month !=x.month)return month<x.month;        else if(day !=x.day)return day<x.day;        else if(hour!=x.hour)return hour < x.hour;        else if(minute != x.minute)return minute < x.minute;    }    int getTicks(){return (day*1440+hour*60+minute);}};int rate[24]={0};int main(){    for(int i=0;i<24;i++)cin>>rate[i];    int cnt=0;    vector<Record> records;    cin>>cnt;    records.reserve(cnt);    for(int i=0;i<cnt;i++){        string _name , _time , _tag;        cin>>_name>>_time>>_tag;        records.push_back(Record(_name , _time , _tag));    }    sort(records.begin() , records.end());    while(records.size()!=0){        vector<Record> tRecords;        double totalCost=0.0;        string selectName=records.front().name;        string seletcMonth=records.front().month;        int index=0;        while(index < records.size() && records[index].name==selectName &&records[index].month==seletcMonth){            tRecords.push_back(records[index]);            index++;        }        records.erase(records.begin() , records.begin() + index);        bool isPrinted=false;        for(int i=0;i<tRecords.size()-1;i++){            if(tRecords[i].tag=="on-line" && tRecords[i+1].tag=="off-line"){                if(isPrinted==false) {cout<<selectName<<" "<<seletcMonth<<endl; isPrinted=true;}                int tick1=tRecords[i].getTicks();                int tick2=tRecords[i+1].getTicks();                int ticks=tick2-tick1;                cout<<tRecords[i].date<<" "<<tRecords[i+1].date<<" "<<ticks<<" $";                double cost=0.0;                if( ticks<=60 && tRecords[i].hour==tRecords[i+1].hour)cost=rate[tRecords[i].hour]*ticks/100.0;                else{                    int min1=60-tRecords[i].minute;                    int min2=tRecords[i+1].minute;                    int totalHours=(ticks-min1-min2)/60;                    cost += min1*rate[tRecords[i].hour]/100.0+min2*rate[tRecords[i+1].hour]/100.0;                    int startHour=tRecords[i].hour+1;                    for(int j=startHour;j<startHour+totalHours;j++)cost+=60.0*rate[ j%24 ]/100.0;                }                totalCost+=cost;                printf("%.2f\n",cost);            }        }        if(totalCost!=0)printf("Total amount: $%.2f\n",totalCost);    }    return 0;}