題目

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

題解

首先判斷是否成對，最後還是用到了map。
成對的話，上線要在下線前。這邊上下線用staus記錄。加在結構體裡。
每組測試資料，所有日期在同一個月，故記錄時間時考慮day,hour,min。

#include<iostream>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
struct customer{
  string name;
  int month,day,hour,min,time,status;//考慮好哪些因素，比如cost和total不在結構體裡
};
bool cmp(customer a,customer b){
  return a.name==b.name?a.time<b.time:a.name<b.name;
}

double count(customer call,int *cost){
  double total=cost[call.hour]*call.min+call.day*cost[24]*60;
  for(int i=0;i<call.hour;i++){
    total+=cost[i]*60;
  }
  
  return total/100.0;
}
int main(){
  int cost[25]={0};
  for(int i=0;i<24;i++){
    scanf("%d",&cost[i]);
    cost[24]+=cost[i];
  }
  int n;
  scanf("%d",&n);
  vector<customer> cus(n);
  for(int i=0;i<n;i++){
    //不能資料型別分開輸入
    cin>>cus[i].name;
    scanf("%02d:%02d:%02d:%02d",&cus[i].month,&cus[i].day,&cus[i].hour,&cus[i].min);
    string temp;
    cin>>temp;
    cus[i].status=(temp=="on-line")?1:0;//這樣可以少用if了
    //因為date的資料都在裡面，可以順便算個通話時間,後面是用掛電話分鐘數-接電話分鐘數
    cus[i].time=cus[i].day*24*60+cus[i].hour*60+cus[i].min;
  }
  sort(cus.begin(),cus.end(),cmp);
  map<string,vector<customer>> result;
  for(int i=1;i<n;i++){
    if(cus[i].name==cus[i-1].name&&cus[i-1].status==1&&cus[i].status==0){
      result[cus[i-1].name].push_back(cus[i-1]);
      result[cus[i].name].push_back(cus[i]);
    }
    
  }
  // 輸出
  for(auto it:result){
   // vector auto::iterator it;
    vector<customer> temp=it.second;//加入map的倒入新的vector遍歷完準備輸出
    //printf(*it.first);
    cout<<it.first;
    printf(" %02d\n",temp[0].month);
    double total=0.0;
    for(int i=1;i<temp.size();i+=2){
      double t=count(temp[i],cost)-count(temp[i-1],cost);
      printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n",temp[i-1].day,temp[i-1].hour,temp[i-1].min,temp[i].day,temp[i].hour,temp[i].min,temp[i].time-temp[i-1].time,t);
      total+=t;
    }
    printf("Total amount: $%.2f\n",total);
    
  }
  return 0;
}

本題最重要的是兩個vector和一個map的運用，map也可以象有序容器一樣push_back。
還有雙重排序，先排名字，再按名字排時間。
這題犯了不少錯。比如輸入輸出格式：%02d老寫成02%d。
最致命的錯就是，當輸入字串時，最好用cin格式，用scanf顯示段錯誤。
多加註意！