1016 Phone Bills (25)（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 n
o 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-lineSample 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: $28.25aaa 0102:00:01 04:23:59 4318 $638.80
Total amount: $638.80将通话记录按照姓名、时间进行排序，进行遍历，当相邻两个记录姓名相同并且上一个记录为on-line，下一个记录为off-line时，才是有效的记录，然后将这些数据 以姓名为键存放到map里。当计算通话费用时，分别计算从第0天第0时第0分到开始时间和结束时间的费用，然后作差。
1 #include <iostream>
2 #include <algorithm>
3 #include <string>
4 #include <map>
5 #include <vector>
6 using namespace std;
7
8 struct Node
9 {
10 string name;
11 int d, h, m, time;
12 bool on_off;
13 };
14
15 int toll[25];
16 bool cmp(Node a, Node b);
17 double billFromZero(Node a);
18
19 int main()
20 {
21 int i, N, month;
22 string s;
23 Node node[1000];
24 for (i = 0; i < 24; i++)
25 {
26 cin >> toll[i];
27 toll[24] += toll[i];
28 }
29 cin >> N;
30 for (i = 0; i < N; i++)
31 {
32 cin >> node[i].name;
33 scanf_s("%d:%d:%d:%d", &month, &node[i].d, &node[i].h, &node[i].m);
34 node[i].time = ((node[i].d * 24) + node[i].h) * 60 + node[i].m;
35 cin >> s;
36 node[i].on_off = (s == "on-line") ? true : false;
37 }
38 sort(node, node + N, cmp);
39 map<string, vector<Node>> mapp;
40 for (i = 1; i < N; i++)
41 {
42 if (node[i].name == node[i - 1].name && !node[i].on_off && node[i - 1].on_off)
43 {
44 mapp[node[i].name].push_back(node[i - 1]);
45 mapp[node[i].name].push_back(node[i]);
46 }
47 }
48 for (auto it : mapp)
49 {
50 vector<Node> v = it.second;
51 cout << it.first;
52 printf(" %02d\n", month);
53 double t, sum = 0.0;
54 for (i = 1; i < v.size(); i += 2)
55 {
56 printf("%02d:%02d:%02d %02d:%02d:%02d %d ", v[i - 1].d, v[i - 1].h, v[i - 1].m, v[i].d, v[i].h, v[i].m, v[i].time - v[i - 1].time);
57 t = billFromZero(v[i]) - billFromZero(v[i - 1]);
58 printf("$%.2f\n", t);
59 sum += t;
60 }
61 printf("Total amount: $%.2f\n", sum);
62 }
63 return 0;
64 }
65
66 bool cmp(Node a, Node b)
67 {
68 return (a.name != b.name) ? (a.name < b.name) : (a.time < b.time);
69 }
70
71 double billFromZero(Node a)
72 {
73 double sum = a.d * toll[24] * 60 + a.m * toll[a.h];
74 for (int i = 0; i < a.h; i++)
75 sum += toll[i] * 60;
76 return sum / 100;
77 }
原文地址：https://www.cnblogs.com/lxc1910/p/9501475.html