1.题目要求把聊天记录分类匹配，然后输出相应的话费和记录。
2.判断记录好坏，把所有记录以string存起来，按照字典序排序，相邻的两条记录满足要求：第一条为on-line且第二条为off-line,才是成功匹配；
3.采用计算00:00:00到现在的耗费和分钟数，这个方法比较简单
4.最重要的：题目中只强调至少有一条匹配成功的记录，不是说每个人都至少有一条匹配成功的记录，所以没有成功匹配记录的人不输出（主要卡在这）
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 $12.10 Total amount: $12.10 CYLL 01 01:06:01 01:08:03 122 $24.40 28:15:41 28:16:05 24 $3.85 Total amount: $28.25 aaa 01 02:00:01 04:23:59 4318 $638.80 Total amount: $638.80
AC代码：
/*
1.判断记录好坏，把所有记录以string存起来，按照字典序排序，相邻的两条记录满足要求：第一条为on-line且第二条为off-line,才是成功匹配；
2.采用计算00:00:00到现在的耗费和分钟数，这个方法比较简单
3.最重要的：题目中只强调至少有一条匹配成功的记录，不是说每个人都至少有一条匹配成功的记录，所以没有成功匹配记录的人不输出（主要卡在这）
*/
//#include<string>
//#include <iomanip>
#include<vector>
#include <algorithm>
//#include<stack>
#include<set>
#include<queue>
#include<map>
//#include<unordered_set>
//#include<unordered_map>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
using namespace std;
struct customNode{
string month;
vector<string> online;
vector<string> offline;
vector<pair<string, string>> recordTag;
vector<pair<string, string>> pairRecord;
vector<pair<int, int>> eachCost;
int total;
customNode() :month(""), online(0), offline(0), eachCost(0), total(0){};
};
int str2int(string a){ return (a[0] - '0') * 10 + a[1] - '0'; }
int calculateCostFromZero(string a, vector<int>& rate)
{//计算从00:00:00到某个时间的耗费
int rateSum = 0;//一天的总和
for (int i = 0; i < 24; i++)
{
rateSum += rate[i] * 60;
}
int day = str2int(a.substr(3, 2));//天数
int hour = str2int(a.substr(6, 2));//小时数
int minute = str2int(a.substr(9, 2));//小时数
int totalCost = day*rateSum;
totalCost += minute*rate[hour];
for (int i = 0; i < hour; i++)
{
totalCost += rate[i] * 60;
}
return totalCost;
}
int calculateMinutesFromZero(string a)
{//计算从00:00:00到某个时间的分钟数
int day = str2int(a.substr(3, 2));//天数
int hour = str2int(a.substr(6, 2));//小时数
int minute = str2int(a.substr(9, 2));//小时数
return day * 1440 + hour * 60 + minute;
}
pair<int, int> calculate(string a, string b, vector<int>& rate)
{//分别计算00:00:00和当前日期的差值
int startCost = calculateCostFromZero(a, rate);
int endCost = calculateCostFromZero(b, rate);
int startMin = calculateMinutesFromZero(a);
int endMin = calculateMinutesFromZero(b);
return{ endCost - startCost, endMin - startMin };
}
pair<int, int> calculate2(string a, string b, vector<int>& rate)
{//计算差值的方法不是很好
int startHour = str2int(a.substr(6, 2));//计算开始所在的小时
int endHour = str2int(b.substr(6, 2));//计算结束时所在的小时
int startNoDay = startHour * 60 + str2int(a.substr(9, 2));//计算H*60+m，单位为分钟
int endNoDay = endHour * 60 + str2int(b.substr(9, 2));//计算H*60+m，单位为分钟
int startWithDay = str2int(a.substr(3, 2)) * 1440 + startNoDay;//加上天数，单位为分钟
int endWithDay = str2int(b.substr(3, 2)) * 1440 + endNoDay;//加上天数，单位为分钟
int days = (endWithDay - startWithDay) / 1440;//查看隔了多少天
int total = 0;
for (int j = 0; j < days; j++)
{//每多一天，都需要加上24小时的计费
for (int i = 0; i < 24; i++)
total += 60 * rate[i];
}
int recent = startNoDay;
if (startNoDay > endNoDay)
{
endHour += 24;//如果起始的分钟数大于结束的分钟数，结束在第一天之后（有可能是2，3，4天），直接+24，便于统计
endNoDay += 24 * 60;//No也需要加上24*60
}
for (int i = startHour + 1; i <= endHour && endHour != startHour; ++i)
{//开始统计
total += (i * 60 - recent)*rate[i - 1];
recent = i * 60;
}
total += (endNoDay - recent)*rate[endHour];
int diff = endWithDay - startWithDay;
return{ total, diff };
}
bool cmpRecordTag(const pair<string, string>&a, const pair<string, string>&b)
{
return a.first < b.first;
}
int main(void)
{
vector<int> rate(48);// = { 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, 10, 10, 10, 10, 10, 20, 20, 20, 15, 15, 15, 15, 15, 15, 15, 20, 30, 20, 15, 15, 10, 10, 10 };
//int rateSum = 0;
//for (int i = 0; i < 24; i++)
//{
// rateSum += rate[i]*60;
//}
//vector<pair<int, int>> temp(0);
//temp.push_back(calculate("01:28:23:41", "01:28:23:42", rate));
//temp.push_back(calculate("01:28:15:41", "01:28:16:05", rate));
//temp.push_back(calculate("01:01:05:59", "01:01:07:00", rate));
//temp.push_back(calculate("01:02:00:01", "01:04:23:59", rate));
//temp.push_back(calculate("01:01:06:01", "01:01:08:03", rate));
//
for (int i = 0; i < 24; i++)
{
scanf("%d", &rate[i]);
rate[i + 24] = rate[i];
}
int recordNum;
cin >> recordNum;
map<string, customNode> m;
vector<string> nameList(0);
for (int i = 0; i < recordNum; i++)
{//输入各条记录
string name, time, tag;
cin >> name >> time >> tag;
if (m.find(name) == m.end())
nameList.push_back(name);
m[name].recordTag.push_back({ time, tag });
}
//直接sort，则string就是按照字典序进行排列
sort(nameList.begin(), nameList.end());
for (int i = 0; i < nameList.size(); i++)
{
string name = nameList[i];
sort(m[name].recordTag.begin(), m[name].recordTag.end(), cmpRecordTag);
m[name].total = 0;
for (int j = 0; j + 1< m[name].recordTag.size(); j++)
{
if (m[name].recordTag[j].second == "on-line"&&m[name].recordTag[j + 1].second == "off-line")
{//当前记录为on-line，且下一条为off-line，则匹配成功
m[name].pairRecord.push_back({ m[name].recordTag[j].first.substr(3), m[name].recordTag[j + 1].first.substr(3) });
m[name].eachCost.push_back(calculate(m[name].recordTag[j].first, m[name].recordTag[j + 1].first, rate));
m[name].total += m[name].eachCost.back().first;
}
}
if (m[name].pairRecord.size() != 0)
{//题目只确保至少有一条成功匹配的记录，但是没有说每个人至少有一条成功匹配的记录
//没成功匹配记录的人就不输出
cout << name << " " << m[name].recordTag[0].first.substr(0, 2) << endl;
for (int j = 0; j < m[name].pairRecord.size(); j++)
{
cout << m[name].pairRecord[j].first << " " << m[name].pairRecord[j].second << " " << m[name].eachCost[j].second << " $";
printf("%.2f", m[name].eachCost[j].first / 100.0);
cout << endl;
}
printf("Total amount: $%.2f", m[name].total / 100.0);
cout << endl;
}
}
return 0;
}