PAT A1016 Phone Bills (25 分) 排序

       题目大意：统计电话费用。一天24小时，给出每个小时内的电话费(cent/min)。之后给出N条通话记录，每条记录包括用户的姓名、记录时间和通话状态(on/off)。要求为每个用户都生成一份账单。值得注意的是， 对同一个用户而言，有效的通话记录是在时间上相邻的两条，前一条是on, 后一条是off。不满足这个条件的通话记录全部忽略。例如，某个用户的记录中， 6h01m 是on，5h59m也是on, 7h0m是off，那么有效的通话时间是 6h01m - 7h0m，因为它们在时间上相邻且状态不同。这一要求与PAT A1095 Cars on Campus 相同。

       核心问题是如何得到每个用户的有效记录。这里需要将所有记录先按照用户姓名排序，再按时间排序。然后满足同一用户、时间相邻、状态不同的两条记录，才能被认为是有效的。之后就是繁琐的细节处理问题。

       因为有天数的存在，计算账单时要考虑跨天，以及每个小时电话费用各不相同，所以直接用分钟为单位计算并不合适，考虑定义结构体记录时间信息。然后定义结构体记录客户信息，其中包括客户的on和off的时间序列。最后根据每个客户的on和off的时间序列，来计算时长和费用等等信息。涉及到时间的问题真的是非常繁琐。

AC代码如下：

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstring>

using namespace std;

int price[24] = {0};

struct timeInfo
{
    int month;
    int day;
    int hour;
    int min;
    timeInfo(int month, int day, int hour, int min): month(month), day(day), hour(hour), min(min){};
    bool operator< (const timeInfo& another)
    {
        if(this->month != another.month) return this->month < another.month;
        else if(this->day != another.day) return this->day < another.day;
        else if(this->hour != another.hour) return this->hour < another.hour;
        else return this->min < another.min;
    }
};

struct record
{
    string name;
    timeInfo time;
    int status;
    record(string name, timeInfo time, int status): name(name), time(time), status(status){};
    bool operator< (const record& another)
    {
        if(this->name != another.name) return this->name < another.name;
        return this->time < another.time;
    }
};

struct customer
{
    string name;
    int id;
    vector<timeInfo> on;
    vector<timeInfo> off;
    customer(string name, int id, timeInfo on, timeInfo off)
    {
        this->name = name;
        this->id = id;
        this->on.push_back(on);
        this->off.push_back(off);
    }
};

map<string, int> nameToId;
int customerCount = 0;
int getId(string name)
{
    if(nameToId.find(name) != nameToId.end()) return nameToId[name];
    nameToId[name] = customerCount;
    return customerCount++;
}

int calCost(const customer& c)
{
    int totalCost = 0;
    for (int i = 0; i < c.on.size(); ++i)
    {
        timeInfo on = c.on[i];
        timeInfo off = c.off[i];
        int totalMin = 0;
        int cost = 0;
        while(on.day != off.day || on.hour != off.hour || on.min != off.min)
        {
            on.min++;
            cost += price[on.hour];
            totalMin++;
            if(on.min == 60)
            {
                on.min = 0;
                on.hour++;
            }
            if(on.hour == 24)
            {
                on.hour = 0;
                on.day++;
            }
        }
        totalCost += cost;
        printf("%02d:%02d:%02d ", c.on[i].day, c.on[i].hour, c.on[i].min);
        printf("%02d:%02d:%02d ", c.off[i].day, c.off[i].hour, c.off[i].min);
        printf("%d $%.2f\n", totalMin, cost * 1.0 / 100);
    }
    printf("Total amount: $%.2f\n", totalCost * 1.0 / 100);
}

int main()
{
    for (int i = 0; i < 24; ++i)
    {
        scanf("%d", &price[i]);
    }
    int N;
    cin >> N;
    vector<record> all;
    for (int i = 0; i < N; ++i)
    {
        char name[30], status[10];
        int month, day, h, m;
        scanf("%s %d:%d:%d:%d %s", name, &month, &day, &h, &m, status);
        all.push_back(record(string(name), timeInfo(month, day, h, m), status[1] == 'n'? 1 : -1));
    }
    sort(all.begin(), all.end());
    int index = 0;
    vector<customer> vCustomer;
    while(index < all.size() - 1)
    {
        if(all[index].name == all[index+1].name && all[index].status == 1 && all[index+1].status == -1)
        {
            int id = getId(all[index].name);
            if(id == vCustomer.size())
            {
                vCustomer.push_back(customer(all[index].name, id, all[index].time, all[index+1].time));
            }
            else
            {
                vCustomer[id].on.push_back(all[index].time);
                vCustomer[id].off.push_back(all[index+1].time);
            }
            index++;
        }
        index++;
    }
    for (int i = 0; i < vCustomer.size(); ++i)
    {
        printf("%s %02d\n", vCustomer[i].name.c_str(), vCustomer[i].on[0].month);
        calCost(vCustomer[i]);
    }
    return 0;
}