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.
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.
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.
给定每个人的通话记录，求每个人的花费。
排序
1、“时间戳”思维，将时间和金额都转化为时间戳的形式；
2、先按照人名和时间排序；
3、之后按照on-line、off-line对数据进行清洗；
4、计算每个人的金额，输出；这里也可以使用map<string, vector<node> >。
五、代码
#define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<algorithm> #include<string> #include<vector> using namespace std; struct node { string name; bool status; int month; int d, h, m; int timestamp; }; int c[24]; int total_num = 0; bool cmp(node n1,node n2) { if (n1.name != n2.name) return n1.name < n2.name; else return n1.timestamp < n2.timestamp; } float getMoney(int d, int h, int m) { double money = 0.0; //day money = d *1.0 * total_num * 60; //hour for (int i = 0; i < h; ++i) { money += 1.0*c[i] * 60; } //minutes money += 1.0 * c[h] * m; return money / 100; } int main() { //read for (int i = 0; i < 24; ++i) { cin >> c[i]; total_num += c[i]; } int n; cin >> n; vector<node> vec; for (int i = 0; i < n; ++i) { node tmp; string s; cin >> tmp.name; scanf("%d:%d:%d:%d", &tmp.month, &tmp.d, &tmp.h, &tmp.m); cin >> s; if (s == "on-line") tmp.status = 0; else tmp.status = 1; //change to timestamp tmp.timestamp = tmp.d * 60* 24 + tmp.h * 60 + tmp.m; vec.push_back(tmp); } //sort sort(vec.begin(), vec.end(), cmp); //clear status int i = 0; vector<node> vec_clear; while (i + 1 < vec.size()) { if (vec[i].name == vec[i + 1].name&&vec[i].status == 0 && vec[i + 1].status == 1) { vec_clear.push_back(vec[i]); vec_clear.push_back(vec[i + 1]); i += 2; } else i++; } //solve every one string s; for (int i = 0; i + 1 < vec_clear.size(); ) { //new body if (vec_clear[i].name != s) { printf("%s %02d\n", vec_clear[i].name.c_str(), vec_clear[i].month); s = vec_clear[i].name; } //every record double total = 0.0; while (i + 1 < vec_clear.size()&&vec_clear[i].name == s) { //get minutes int min = vec_clear[i + 1].timestamp - vec_clear[i].timestamp; //get money double money = 0.0; money = getMoney(vec_clear[i + 1].d, vec_clear[i + 1].h, vec_clear[i + 1].m) - getMoney(vec_clear[i].d, vec_clear[i].h, vec_clear[i].m); total += money; printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2lf\n", vec_clear[i].d, vec_clear[i].h, vec_clear[i].m, vec_clear[i + 1].d, vec_clear[i + 1].h, vec_clear[i + 1].m, min,money); i += 2; } printf("Total amount: $%.2lf\n", total); } system("pause"); return 0; }