1016 Phone Bills （25 分)

2019-03-10

1016 Phone Bills （25 分)

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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出24小时电话费以及通话记录（时间，上下线），求各人的电话费。注意必须是上线之后接着下线才是有效数据。电话费为0的账单不输出。

分析

把个人的记录按时间排序，依次处理。只有上线之后紧接着的记录为下线的是有效记录,为其生成一条账单。这里计算电话费的方式比较巧妙，参考了柳婼の blog

的处理方法。计算从0点开始的至上线及下线时刻的话费，两个相减即为实际话费，避免了繁琐的逻辑。

代码（未AC，差测试点3未通过）

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <cstring>
#include <cmath>

using namespace std;

struct record{
    string name;
    int mon;
    int day;
    int hour;
    int mm;
    int t;
    string on;

    record(const string &name, int mon, int day, int hour, int mm, const string &on) : name(name), mon(mon), day(day),
                                                                                       hour(hour), mm(mm), on(on) {t=day*1440+hour*60+mm;}
};
double feday=0;
int fee[24];
struct bill{
    int d1,d2;
    int h1,h2;
    int m1,m2;
    int t;
    double f;
};
bool cmp(record r1,record r2){
    if(r1.t<r2.t)
        return true;
    return false;
}
double charge(record r1,record r2){

    double f1=(r1.day-1)*feday;
    double f2=(r2.day-1)*feday;
    for(int i=0;i<r1.hour;i++)
        f1+=60*fee[i];
    for(int i=0;i<r2.hour;i++)
        f2+=60*fee[i];
    f1+=r1.mm*fee[r1.hour];
    f2+=r2.mm*fee[r2.hour];
    return (f2-f1)/100;
}
int main() {
    string name,on;
    vector<string> names;
    int mon,day,hour,mm;
    map<string,vector<bill>> bills;
    map<string,vector<record>> maps;
    for(int i=0;i<24;i++) {
        cin >> fee[i];
        feday+=fee[i]*60;
    }
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>name;
        scanf("%d:%d:%d:%d",&mon,&day,&hour,&mm);
        cin>>on;
        if(find(names.begin(),names.end(),name)==names.end())
            names.push_back(name);
        record r(name,mon,day,hour,mm,on);
        maps[name].push_back(r);
    }
    sort(names.begin(),names.end());
    for(string name:names){
        vector<record> vecname=maps[name];
        sort(vecname.begin(),vecname.end(),cmp);
        for(auto it=vecname.begin()+1;it!=vecname.end();it++){
            if(it->on=="off-line"&&(it-1)->on=="on-line"){
                bill nb;
                nb.d2=it->day;
                nb.d1=(it-1)->day;
                nb.h2=it->hour;
                nb.h1=(it-1)->hour;
                nb.m2=it->mm;
                nb.m1=(it-1)->mm;
                nb.f=charge(*(it-1),*it);
                nb.t=it->t-(it-1)->t;
                bills[name].push_back(nb);
            }
        }
    }
    for(string name:names){
        double ff=0;

        vector<bill> v=bills[name];
        for(bill b:v)
            ff+=b.f;

        if(ff>=0&&!v.empty()) {
            cout << name;
            printf(" %02d\n", mon);

                for (bill b:v)
                    printf("%02d:%02d:%02d %02d:%02d:%02d %02d $%.2f\n", b.d1, b.h1, b.m1, b.d2, b.h2, b.m2, b.t, b.f);
                printf("Total amount: $%.2f\n", ff);

        }
    }
}