PAT Advanced 1017. Queueing at B

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题目

Suppose a bank has windows open for service. There is a yellow line in
front of the windows which devides the waiting area into two parts. All the
customers have to wait in line behind the yellow line, until it is his/her
turn to be served and there is a window available. It is assumed that no
window can be occupied by a single customer for more than 1 hour.

Now given the arriving time and the processing time of each customer,
you are supposed to tell the average waiting time of all the customers.

Input Specification:

Each input file contains one test case. For each case, the first line contains
2 numbers: ( ) - the total number of customers, and ( ) - the number of windows. Then lines follow, each contains 2 times:
HH:MM:SS - the arriving time, and - the processing time in minutes of a
customer. Here HH is in the range [00, 23], MM and SS are both in [00,
59]. It is assumed that no two customers arrives at the same time.

Notice that the bank opens from 08:00 to 17:00. Anyone arrives early will have
to wait in line till 08:00, and anyone comes too late (at or after 17:00:01)
will not be served nor counted into the average.

Output Specification:

For each test case, print in one line the average waiting time of all the
customers, in minutes and accurate up to 1 decimal place.

Sample Input:

7 3
07:55:00 16
17:00:01 2
07:59:59 15
08:01:00 60
08:00:00 30
08:00:02 2
08:03:00 10

Sample Output:

8.2

思路

这道题虽然和1014很像, 但是实际上要简单得多. 因为1014题需要具体模拟队列的出入操作,
但是这道题则不需要队列这样的数据结构.

需要注意的点:

• 晚于下午五点的顾客才会不予处理, 之前到来的不管等到多晚, 都会处理
• 处理时长长于一小时需降为一小时

数据结构:

• struct[N]: 记录顾客时间信息, 结构体包含顾客到达时间和处理时长两个整形
• int[K]: 记录每个窗口的顾客完成时间信息

大体思路:

• 读取顾客时间信息, 按照到达时间排序, 以便处理
• 遍历所有到达时间不晚于17:00的顾客:
  • 找到完成时间最早的窗口
  • 如果窗口完成时间晚于当前顾客到达时间
    • 两者之差累加到等待时间
    • 窗口时间累加上顾客处理时长
  • 否则无需等待, 窗口时间更新为顾客到达时间加顾客处理时长

代码

最新代码@github，欢迎交流

#include <stdio.h>
#include <stdlib.h>

typedef struct Customer{
    int start, len;
}Customer;

/* Compare two Customer stucture by start time */
int cmp(const void *a, const void *b)
{
    Customer c1 = *(Customer*)a;
    Customer c2 = *(Customer*)b;
    return c1.start - c2.start;
}

int main()
{
    int N, K, earliest, i;
    int HH, MM, SS;
    int wait_time = 0, queue_time[100] = {0};
    Customer customers[10000], *p;

    scanf("%d %d", &N, &K);
    for(int i = 0; i < N; i++)
    {
        scanf("%d:%d:%d %d", &HH, &MM, &SS, &customers[i].len);
        /* Relative time to 08:00:00 */
        customers[i].start = SS + 60 * (MM + 60 * (HH - 8));
        customers[i].len *= 60;
    }

    qsort(customers, N, sizeof(Customer), cmp);

    for(i = 0; i < N; i++)
    {
        p = customers + i;

        /* Find the queue number which will finish next */
        earliest = 0;
        for(int i = 0; i < K; i++)
            if(queue_time[i] < queue_time[earliest])
                earliest = i;

        /* later than 17:00:00 */
        if(p->start > (17 - 8) * 3600)
            break;
        /* processing time longer than one hour */
        if(p->len > 3600)
            p->len = 3600;

        /* increase total waiting time and modify the time of each queue */
        if(p->start < queue_time[earliest])
        {
            wait_time += queue_time[earliest] - p->start;
            queue_time[earliest] += p->len;
        }
        else
        {
            queue_time[earliest] = p->start + p->len;
        }
    }

    if(i)
        printf("%.1f", wait_time / 60.0 / i);
    else
        printf("0.0");

    return 0;
}