PAT_1016: 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

备注：用了map，把人名当key，此人的records数组作为value，避免了sort人名，因为在map中是按照key自动排序的（string类型刚好是按字典升序排的）。 然后把每个人的records根据时间sort一下，找到匹配的records算一下时间和费用就可以（算费用的过程有够麻烦的）。还要注意的是如果某人没有匹配的记录则不用输出任何东西！（题目中只保证了所有输入中肯定有一对匹配的记录，不保证每个人都有一对匹配的记录）另外，刚开始一直没A居然是因为人名数组开太小了，char name[20]改成char name[50]就对了，我真是天雷滚滚，口吐无限脏话！

#include<stdio.h>#include<iostream>#include<algorithm>#include<map>#include<vector>#include<string>using namespace std;#define ONLINE 1#define OFFLINE -1int toll[24];int n;typedef struct record{char name[50];int month;int day;int hour;int minute;int b_line;}RECORD;map<string, vector<RECORD>> myMap;map<string, vector<RECORD>>::iterator iter;//compare two records based on time//returns true if r1 is earlier than r2//otherwise returns falsebool compareRecords(RECORD r1, RECORD r2){if(r1.month<r2.month)return true;else if(r1.month>r2.month)return false;else{if(r1.day<r2.day)return true;else if(r1.day>r2.day)return false;else{if(r1.hour<r2.hour)return true;else if(r1.hour>r2.hour)return false;else{if(r1.minute<r2.minute)return true;elsereturn false;}}}}//compute the toll for a call, complicated!float ComputeToll(RECORD r1,RECORD r2){int charge = 0;int totalmins = 0;int one_day_charge = 0;for(int i=0;i<24;i++)one_day_charge+=toll[i]*60;// because tolls are computed based on hour, we first compare hourif(r2.hour>r1.hour){for(int i=r1.hour+1;i<r2.hour;i++){charge+= toll[i]*60;totalmins+=60;}charge+=(60-r1.minute)*toll[r1.hour]+r2.minute*toll[r2.hour];totalmins+=(60-r1.minute)+r2.minute;charge+= (r2.day-r1.day)*one_day_charge;totalmins+=(r2.day-r1.day)*60*24;}else if(r2.hour==r1.hour){charge+=(r2.minute-r1.minute)*toll[r1.hour];totalmins+=(r2.minute-r1.minute);charge+= (r2.day-r1.day)*one_day_charge;totalmins+=(r2.day-r1.day)*60*24;}else // r2.hour<r1.hour, two different days, r2.day>r1.day{charge+=(60-r1.minute)*toll[r1.hour];totalmins+=(60-r1.minute);for(int i=r1.hour+1;i<24;i++){charge+= toll[i]*60;totalmins+=60;}for(int i=0;i<r2.hour;i++){charge+= toll[i]*60;totalmins+=60;}charge+=r2.minute*toll[r2.hour];totalmins+=r2.minute;charge+=(r2.day-r1.day-1)*one_day_charge;totalmins+=(r2.day-r1.day-1)*60*24;}printf("%d $%.2f\n",totalmins,(float)charge/(float)100);return (float)charge/(float)100;}// pair the records for one personvoid PairRecords(vector<RECORD> r_list){// sort the records by timesort(r_list.begin(),r_list.end(),compareRecords);vector<RECORD>::iterator v_iter;bool havePairs = false;float amount = 0;for(v_iter = r_list.begin();v_iter != r_list.end(); v_iter++){RECORD r = *v_iter;if(r.b_line == ONLINE){if((v_iter+1)!=r_list.end()) {RECORD r_next = *(v_iter+1);if(r_next.b_line == OFFLINE) // find a pair{if(!havePairs) // note that a customer could have no paired records, then no output should be given{printf("%s %02d\n",r.name, r.month);havePairs = true;}//compute the toll of the current callprintf("%02d:%02d:%02d %02d:%02d:%02d ",r.day,r.hour,r.minute,r_next.day,r_next.hour,r_next.minute);float charge = ComputeToll(r,r_next);amount+=charge;}}}}if(havePairs)printf("Total amount: $%.2f\n",amount);}int main(){for(int i=0;i<24;i++)cin>>toll[i];cin>>n;for(int i=0;i<n;i++){RECORD r;cin>>r.name;scanf("%d:%d:%d:%d",&r.month,&r.day,&r.hour,&r.minute);string line;cin>>line;if(line=="on-line")r.b_line = ONLINE;elser.b_line = OFFLINE;myMap[r.name].push_back(r);}// pair the record for each personfor(iter = myMap.begin();iter != myMap.end(); iter++){vector<RECORD> r_list = iter->second;PairRecords(r_list);}return 0;}