PAT A1016 Phone Bills （25 分）

Posted 2021-02-09 tccbj

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 (≤), 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

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
#include <string>
#include <set>
#include <map>
using namespace std;
const int maxn = 1010;
int n;
string name,line;
int m, d, h, mm, all = 0;
int cost[24] = { 0 };
struct person {
    string name;
    int time;
    int d;
    int h;
    int mm;
    int status;
};
bool cmp(person p1, person p2) {
    return p1.name == p2.name ? p1.time < p2.time : p1.name < p2.name;
}
float cal_cost(person p) {
    float c = 0;
    c += p.d * all * 60;
    for (int j = 0; j < p.h; j++) {
        c += cost[j] * 60;
    }
    c += cost[p.h] * p.mm;
    return c;
}
vector<person> v;
map<string, vector<person> > mp;
int main() {
    for (int i = 0; i < 24; i++) {
        scanf("%d", &cost[i]);
        all += cost[i];
    }
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        cin >> name;
        scanf("%d:%d:%d:%d", &m, &d, &h, &mm);
        cin >> line;
        person p;
        p.name = name;
        if (line == "on-line") {
            p.status = 0;
        }
        else {
            p.status = 1;
        }
        p.d = d;
        p.h = h;
        p.mm = mm;
        p.time = 1440 * d + 60 * h + mm;
        v.push_back(p);
    }
    sort(v.begin(), v.end(), cmp);
    for (int i = 1; i < n; i++) {
        if (v[i].name == v[i - 1].name && (v[i].status == 1 && v[i - 1].status == 0)) {
            mp[v[i].name].push_back(v[i - 1]);
            mp[v[i].name].push_back(v[i]);
        }
    }
    for (auto it = mp.begin(); it != mp.end(); it++) {
        string name = it->first;
        cout << name;
        printf(" %02d
", m);
        float total = 0;
        vector<person> pv = it->second;
        for (int i = 1; i < pv.size(); i += 2) {
            float cost_1 = 0, cost_2 = 0;
            cost_1 = cal_cost(pv[i - 1]);
            cost_2 = cal_cost(pv[i]);
            float total_cost = (cost_2 - cost_1) / 100;
            total += total_cost;
            printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f
", pv[i - 1].d, pv[i - 1].h, pv[i - 1].mm, pv[i].d, pv[i].h, pv[i].mm, pv[i].time - pv[i - 1].time, total_cost);
        }
        printf("Total amount: $%.2f
", total);
    }
    system("pause");
    return 0;
}

注意点：逻辑很简单的一道题，我却做了4个小时多才AC，实在太菜了。

一开始没想到这道题本质是个排序题，看到要按从小到大排列，就想到了set和map自动排序，然后就进去出不来了，一直纠结在怎么判断两个通话记录是一对上。

第二个坑是计算花费，看了大佬的思路才发现原来可以通过相对值的差来得到，而不一定非要把两个时间的差求出来，把每个时间点相对0的价格算出来以后的差就是所要的价格。

第三点是最后输出结果，对每个人还要输出一个总和，自己只想到了用各种if判断，看了大佬思路，用map的vector实在是太方便了，而且还准确。