A1016 Phone Bills (25)（25 分）

A1016 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 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

思路

这里面时间轴的思想非常重要。

日期和时间的排序，以及，分钟数计算时间差的手段

K&R的书里竟然有错，在strcmp函数介绍里，不过总算是明白cmp函数返回值得原理。

升序是a>b返回正值

降序是a<b返回正值

分钟计时的代码也很不错

while(temp.dd < rec[off].dd || temp.hh < rec[off].hh || temp.mm < rec[off].mm) {
        (*time)++;
        (*money)+= toll[temp.hh];
        temp.mm++;
        if(temp.mm >= 60) {
            temp.mm = 0;
            temp.hh++;
        }
        if(temp.hh >= 24) {
            temp.hh = 0;
            temp.dd++;
        }
    }

AC代码

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
#define maxn  1010
int toll[25];
struct Record {
    char name[25];
    int month, dd, hh, mm;
    bool status;
} rec[maxn], temp;
int cmp(const void* a, const void* b) {
	struct Record*aa=(Record*)a;
	struct Record*bb=(Record*)b;
//	struct Record*aa=a;
//	struct Record*bb=b;

    int s = strcmp(aa->name, bb->name);
/*	aa->name<bb->name时，返回负值，aa->name=bb->name时，返回0 ，aa->name>bb->name时，返回正值
    if(s != 0) return s > 0;*/
/*	为使qsort达成升序排列
	在函数cmp中，如果第一个参数小于第二个参数，它必须返回一个负值；
	如果第一个参数等于第二个参数，它必须返回0；
	如果第一个参数大于第二个参数，它必须返回一个正值。
	*/

    if(s != 0) return s>0 ? 1:-1;/*客户的名字，字典序，升序*/
    else if(aa->month != bb->month) return aa->month > bb->month;/*月份升序*/
    else if(aa->dd != bb->dd) return aa->dd > bb->dd;/*日期升序*/
    else if(aa->hh != bb->hh) return aa->hh > bb->hh;/*钟点升序*/
    else return aa->mm > bb->mm;/*分钟升序,数据限制，没有到分钟都全部相同的记录*/
}
void get_ans(int on, int off, int* time, int* money) {
    temp = rec[on];
    while(temp.dd < rec[off].dd || temp.hh < rec[off].hh || temp.mm < rec[off].mm) {
        (*time)++;
        (*money)+= toll[temp.hh];
        temp.mm++;
        if(temp.mm >= 60) {
            temp.mm = 0;
            temp.hh++;
        }
        if(temp.hh >= 24) {
            temp.hh = 0;
            temp.dd++;
        }
    }
}
int main() {
    for(int i = 0; i < 24; i++) {
        scanf("%d", &toll[i]);
    }
    int n;
    scanf("%d", &n);
    char line[10];
    for(int i = 0; i < n; i++) {
        scanf("%s", rec[i].name);
        scanf("%d:%d:%d:%d", &rec[i].month, &rec[i].dd, &rec[i].hh, &rec[i].mm);
        scanf("%s", line);
        if(strcmp(line, "on-line") == 0) {/*字符串相等*/
            rec[i].status = true;
        } else {
            rec[i].status = false;
        }
    }
    qsort(rec, n, sizeof (struct Record), cmp);/*排序有问题*/
//    for(int i=0;i<n;i++){
//    	printf("%s ",rec[i].name);
//    	printf("%d:%d:%d:%d\n",rec[i].month,rec[i].dd,rec[i].hh,rec[i].mm);
//    //	printf("%s",rec[i].status);
//
//	}
    int on = 0, off, next;//next下一位客户
    while(on < n) {
        int needPrint = 0;
        next = on;
        while(next < n && strcmp(rec[next].name, rec[on].name) == 0) {
            if(needPrint == 0 && rec[next].status == true) {
                needPrint = 1;
            } else if(needPrint == 1 && rec[next].status == false) {
                needPrint = 2;
            }
            next++;
        }/*在同一客户记录中寻找有无匹配的时间轴记录*/
        if(needPrint < 2) {/*没有成对的，时间轴相邻的on-line 与off-line*/
            on = next;
            continue;/*跳过这些记录*/
        }
        int AllMoney = 0;
        printf("%s %02d\n", rec[on].name, rec[on].month);/*该用户存在有效通话记录*/
        while(on < next) {
            while(on < next - 1
                  && !(rec[on].status == true && rec[on + 1].status == false)) {
                on++;/*直到找到连续的on-line和off-line*/
            }
            off = on + 1;
            if(off == next) {
                on = next;
                break;
            }
            printf("%02d:%02d:%02d ", rec[on].dd, rec[on].hh, rec[on].mm);
            printf("%02d:%02d:%02d ", rec[off].dd, rec[off].hh, rec[off].mm);
            int time = 0, money = 0;
            get_ans(on, off, &time, &money);
            AllMoney += money;
            printf("%d $%.2f\n", time, money / 100.0);
            on = off + 1;
        }
        printf("Total amount: $%.2f\n", AllMoney / 100.0);
    }
    return 0;
}