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 10
10
CYLL 01:01:06:01 on-line
CYLL 01:28:16:05 off-line
CYJJ 01:01:07:00 off-line
CYLL 01:01:08:03 off-line
CYJJ 01:01:05:59 on-line
aaa 01:01:01:03 on-line
aaa 01:02:00:01 on-line
CYLL 01:28:15:41 on-line
aaa 01:05:02:24 on-line
aaa 01:04:23:59 off-line
Sample Output:
CYJJ 01
01:05:59 01:07:00 61 12.10Totalamount:12.10
CYLL 01
01:06:01 01:08:03 122 24.4028:15:4128:16:05243.85
Total amount: 28.25aaa0102:00:0104:23:594318638.80
Total amount: $638.80

解题思路：将所有的记录进行排序，先按名字的字母序排，相同之后时间从小打到排，这样同一个人的所有的记录都是在一起的方便处理。然后将同一个人的所有记录提取出来，进行配对计算输出，配对原则是按照距离最近的on-off配对。
题目坑爹的一点是说每个输入一定至少有一对配对的，但是并不是每个人的记录都一zing至少有一对配对的，对于某些一对配对都没有的顾客不用输出。
最后还是吐槽一下傻逼的自己，cmp函数居然都能写错
return r1.minute < r2.minute;写成return r1.minute < r1.minute; 最后发现错误的我差点哭了。

#include<iostream>#include<stdio.h>#include<vector>#include<algorithm>#include<string.h>using namespace std;struct record{    char name[22];    int month;    int day;    int hour;    int minute;    char status[10];};int tolls[24];bool cmp(const record & r1, const record & r2){    if (strcmp(r1.name, r2.name) < 0){        return true;    }    else if (strcmp(r1.name, r2.name) == 0){        if (r1.day < r2.day){            return true;        }        else if (r1.day == r2.day){            if (r1.hour < r2.hour){                return true;            }            else if (r1.hour == r2.hour){                return r1.minute < r2.minute;            }            else{                return false;            }        }        else{            return false;        }    }    else{        return false;    }}double show_record(record start, record end){    record s = start;    double sum = 0;    int count = 0;    /*while (start.day < end.day || start.hour < end.hour || start.minute < end.minute){        start.minute++;        count++;        sum += tolls[start.hour] * 1.0 / 100.0;        if (start.minute == 60){            start.minute = 0;            start.hour++;        }        if (start.hour == 24){            start.hour = 0;            start.day++;        }    }*/    int count_start = start.minute + start.hour * 60 + start.day * 24 * 60;    int count_end = end.minute + end.hour * 60 + end.day*24*60;    int cost_start = start.minute * tolls[start.hour];    int cost_end = end.minute * tolls[end.hour];    for (int i = 0; i < start.hour; i++){        cost_start += tolls[i] * 60;    }    for (int i = 0; i < end.hour; i++){        cost_end += tolls[i] * 60;    }    int temp = 0;    for (int i = 0; i < 24; i++){        temp += tolls[i];    }    cost_start += start.day*temp * 60;    cost_end += end.day*temp * 60;    sum = (cost_end - cost_start)*1.0/100.0;    printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2lf\n", s.day, s.hour, s.minute, end.day, end.hour, end.minute, count_end-count_start, (cost_end-cost_start)*1.0/100.0);    return sum;}void show(vector<record>vec){    record start;    record end;    double total = 0.0;    vector<record>haha;    for (int i = 0; i < vec.size(); i++){        if (strcmp(vec[i].status, "on-line") == 0){            start = vec[i];            for (i = i + 1; i < vec.size(); i++){                if (strcmp(vec[i].status, "on-line") == 0){                    start = vec[i];                }                if (strcmp(vec[i].status, "off-line") == 0){                    end = vec[i];                    haha.push_back(start);                    haha.push_back(end);                    break;                }            }        }    }    if (haha.size()){        printf("%s %02d\n", haha[0].name, haha[0].month);        for (int i = 0; i < haha.size(); i += 2){            total += show_record(haha[i], haha[i + 1]);        }        printf("Total amount: $%.2lf\n", total);    }}int main(){    for (; scanf("%d", &tolls[0]) != EOF;){        for (int i = 1; i < 24; i++){            scanf("%d", &tolls[i]);        }        int n;        scanf("%d", &n);        vector<record>vec(n);        for (int i = 0; i < n; i++){            scanf("%s %d:%d:%d:%d %s", vec[i].name, &vec[i].month, &vec[i].day, &vec[i].hour, &vec[i].minute, vec[i].status);        }        sort(vec.begin(), vec.end(), cmp);        record start;        record end;        char *name = vec[0].name;        vector<record>out;        out.push_back(vec[0]);        for (int i = 1; i < n; i++){            if (strcmp(name, vec[i].name) == 0){                out.push_back(vec[i]);            }            else{                name = vec[i].name;                show(out);                out.clear();                out.push_back(vec[i]);            }        }        if (out.size()){            show(out);        }    }    return 0;}