1016. Phone Bills (25)

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'

/*典型的排序算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
struct Records
{
	int start, dd, hh, mm;
	bool isOn;
	bool operator < (const Records &rec) const
	{
		return start < rec.start;
	}
};
struct Call
{
	int sd, sh, sm, ed, eh, em, time;
	double money;
};
vector<Records> records[500];
vector<Call> calls[500];
map<string, int> mp;
int main(void)
{
	//
	int rate[24]; //不同的小时内每分钟收取的费用
	for (int i = 0; i < 24; ++i)
		scanf("%d", &rate[i]);
	//
	int n;
	scanf("%d", &n);
	char name[21], status[10];
	int mon;//月份
	int count = 0, index;
	map<string, int> :: iterator iter;
	Records rec;
	for (int i = 0; i < n; ++i)
	{
		//输入电话记录，name, 
		scanf("%s%d:%d:%d:%d%s", name, &mon, &rec.dd, &rec.hh, &rec.mm, status);
		iter = mp.find(name); //给每个name分配一个下标
		if (iter == mp.end())
		{
			mp.insert(make_pair(name, count));
			index = count++; //count标志有多少个用户, index标志name映射的下标
		}
		else
			index = iter -> second;
		if (status[1] == 'n')
			rec.isOn = true;
		else
			rec.isOn = false;
		rec.start = rec.dd * 24 * 60 + rec.hh * 60 + rec.mm;
		records[index].push_back(rec);
	}
	int dayMoney = 0, k;
	for (k = 0; k < 24; ++k)
		dayMoney += rate[k] * 60;
	Call call;
	for (int j = 0; j < count; ++j)
	{
		sort(records[j].begin(), records[j].end());
		for (int i = 1; i < records[j].size(); ++i)
		{
			if (records[j][i].isOn == false && records[j][i - 1].isOn == true)
			{
				call.sd = records[j][i-1].dd;
				call.sh = records[j][i-1].hh;
				call.sm = records[j][i-1].mm;
				call.ed = records[j][i].dd;
				call.eh = records[j][i].hh;
				call.em = records[j][i].mm;
				call.time = records[j][i].start - records[j][i-1].start;
				if (call.ed > call.sd)
				{
					
					call.money = (call.ed - call.sd - 1) * dayMoney;
					call.money += (60 - call.sm) * rate[call.sh];
					k = call.sh + 1;
					while (k < 24)
						call.money += 60 * rate[k++];
					k = 0;
					while (k < call.eh)
						call.money += 60 * rate[k++];
					call.money += call.em * rate[call.eh];
				}
				else
				{
					if (call.sh < call.eh)
					{
						call.money = (60 - call.sm) * rate[call.sh];
						k = call.sh + 1;
						while (k < call.eh)
						{
							call.money += 60 * rate[k++];
						}
						call.money += call.em * rate[call.eh];
					}
					else
						call.money = (call.em - call.sm) * rate[call.sh];
				}
				calls[j].push_back(call);
				++i;
			}
			
		}

	}
	for (map<string, int> :: iterator iter = mp.begin(); iter != mp.end(); ++iter)
	{
		if (calls[iter->second].size() == 0)//有有效记录才会有输出，注意如果没有这句代码会导致两个测试案例通不过
			continue;
		printf("%s %02d\n", iter -> first.c_str(), mon);
		int money = 0;
		for (vector<Call> :: iterator vIter = calls[iter->second].begin(); vIter != calls[iter->second].end(); ++vIter)
		{
			printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n", vIter->sd, vIter->sh, vIter->sm, vIter->ed, vIter->eh, vIter->em, vIter->time, vIter->money * 0.01);
			money += vIter -> money;
		}
		printf("Total amount: $%.2f\n", money * 0.01);
	}
}
//改进过后的方法，只有用一个vector保存数据，进行一次排序
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;

struct Call
{
	char str[24];
	int month, day, hour, miniter, total;
	bool On;
	bool operator < (const Call & ca) const
	{
		int legal = strcmp(str, ca.str);
		if ( legal != 0)
			return legal < 0 ? true : false;
		else
			return total < ca.total;
	}
};

vector<Call> call;
int rate[24];
int oneDay;
int oneHour[24];
void show(Call a, Call b)
{
	printf("%02d:%02d:%02d %02d:%02d:%02d %d", a.day, a.hour, a.miniter, b.day, b.hour, b.miniter, b.total - a.total);
}
int getMoney(Call call)
{
	int money = 0;
	for (int i = 0; i < call.day; ++i)
		money += oneDay;
	for (int i = 0; i < call.hour; ++i)
		money += oneHour[i];
	money += call.miniter * rate[call.hour];
	return money;
}
int main(void)
{
	oneDay = 0;
	for (int i = 0; i < 24; ++i)
	{
		scanf("%d", &rate[i]);
		oneHour[i] = rate[i] * 60;
		oneDay += oneHour[i];
	}
	int n;
	scanf("%d", &n);
	call.resize(n);
	char status[10];
	for (int i = 0; i < n; ++i)
	{
		scanf("%s%d:%d:%d:%d%s", call[i].str, &call[i].month, &call[i].day, &call[i].hour, &call[i].miniter, status);
		call[i].total = call[i].day * 24 * 60 + call[i].hour * 60 + call[i].miniter;
		if (status[1] == 'n')
			call[i].On = true;
		else
			call[i].On = false;
	}
	sort(call.begin(), call.end());
	bool hasP = false;
	int money = 0, moneyC;
	for (int i = 1; i < n; ++i)
	{
		if (strcmp(call[i].str, call[i-1].str) != 0)
		{
			if (hasP)
				printf("Total amount: $%.2f\n", money * 0.01);
			hasP = false;
			money = 0;
			continue;
		}
		if (call[i-1].On && !call[i].On) //有效记录
		{
			if (hasP == false)
			{	
				printf("%s %02d\n", call[i].str, call[i].month);
				hasP = true;
			}
			show(call[i-1], call[i]);
			moneyC= getMoney(call[i]) - getMoney(call[i-1]);
			printf(" $%.2f\n", moneyC * 0.01);
			money += moneyC;
		}
	}
	if (hasP)
		printf("Total amount: $%.2f\n", money * 0.01);
	return 0;
}