1016 Phone Bills（模拟）

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

这道题写的有点变形了，要注意的点比较多，有问题的点如下

1.该月没有消费的用户不用打印订单（即需要筛选出所有该月消费不为0的用户再一起输出，而不能边筛选边输出），没意识到这一点，一直挂测试点

2.计算时间差问题，我是自己硬推了一套计算方式，只能过第一个和最后一个测试点，另外两种ac的做法我也注释在了代码里

3.全程一直缝缝补补，数据结构开的乱七八糟，可以优化的点有很多

#include<bits/stdc++.h>
using namespace std;
int toll[25];
int n;
int all_day[26]={0}; //all_day[i]表示从0点到i点的总单价
typedef struct info{
    int month;
    multimap<int,bool> record;
    map<int,int> valid;
};
map<string,info> customer;
map<string,info> result;
int main(){
    for (int i=0;i<24;++i){
        cin>>toll[i];
        all_day[i+1]=all_day[i]+toll[i];
    }
    cin>>n;
    string name,state;
    int month,day,hour,minute;
    for (int i=0;i<n;++i){
        cin>>name;
        scanf("%d:%d:%d:%d",&month,&day,&hour,&minute);
        cin>>state;
        customer[name].month=month;
        customer[name].record.insert(pair<int,bool>(day*24*60+hour*60+minute,state=="on-line"?false:true));
    }
    map<string,info>::iterator c_it;
    multimap<int,bool>::iterator r_it;
    map<int,int>::iterator i_it;
    for (c_it=customer.begin();c_it!=customer.end();++c_it){ //筛选有效通话记录
        multimap<int,bool> temp=c_it->second.record;
        int on=-1;
        for (r_it=temp.begin();r_it!=temp.end();++r_it){
            if (r_it->second){  //该条记录为off-line状态
                if (on>=0){ //之前存在on-line状态，可以结算该次通话费用
                    result[c_it->first].valid[on]=r_it->first;
                    result[c_it->first].month=c_it->second.month;
                    on=-1;
                }
            }
            else
                on=r_it->first;
        }
    }
    for (c_it=result.begin();c_it!=result.end();++c_it){
        cout<<c_it->first<<" ";
        printf("%02d\n",c_it->second.month);
        map<int,int> temp=c_it->second.valid;
        double sum=0;
        for (i_it=temp.begin();i_it!=temp.end();++i_it){
            int d1,d2,h1,h2,m1,m2;
            d1=i_it->first/1440;d2=i_it->second/1440;
            h1=(i_it->first-d1*1440)/60;h2=(i_it->second-d2*1440)/60;
            m1=i_it->first-d1*1440-h1*60;m2=i_it->second-d2*1440-h2*60;
            printf("%02d:%02d:%02d %02d:%02d:%02d %d",d1,h1,m1,d2,h2,m2,i_it->second-i_it->first);
            //计算费用
            double bill=0;

            //从月初开始计费，则后一时间减前一时间就是通话时间
            double bill1=0,bill2=0;
            bill1+=all_day[24]*d1*60+all_day[h1]*60+toll[h1]*m1;
            bill2+=all_day[24]*d2*60+all_day[h2]*60+toll[h2]*m2;
            bill=bill2-bill1;

            //直接模拟法
            /*while(d1!=d2||h1!=h2||m1!=m2){ 
                bill+=toll[h1];
                m1++;
                if (m1==60){
                    m1=0;
                    h1++;
                }
                if (h1==24){
                    h1=0;
                    d1++;
                }
            }*/

            //自己硬推的计算法
            /*if (d2>d1){ 
                if (h2>h1)//若经过整日
                    bill+=(d2-d1)*all_day[24]*60;
                else if (h2<h1)
                    bill+=(d2-d1-1)*all_day[24]*60;
                else
                    if (m2>=m1)
                        bill+=(d2-d1)*all_day[24]*60;
                    else
                        bill+=(d2-d1-1)*all_day[24]*60;
            } 
            if (all_day[h2]>all_day[h1]){ //同天
                bill+=(all_day[h2]-all_day[h1+1])*60;
                bill+=toll[h1]*(60-m1)+toll[h2]*m2;
            }
            else if (all_day[h2]<all_day[h1]){ //隔天
                bill+=(all_day[24]-all_day[h1]+all_day[h2+1])*60;
                bill-=toll[h1]*m1+toll[h2]*(60-m2);
            }
            else //同时
                bill+=toll[h1]*(m2-m1);*/

            bill/=100;
            sum+=bill;
            printf(" $%.2f\n",bill);
        }
        printf("Total amount: $%.2f\n",sum);
    }
}