PAT 甲级 1017. Queueing at Bank

本文介绍了一个银行排队系统的模拟算法,该算法通过计算不同客户的到达时间和处理时间,来预测每个客户的平均等待时间。采用C++实现,包括使用结构体存储客户信息、对客户按到达时间排序及动态分配窗口处理客户请求等步骤。

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原题传送门

  • 题意:有N个客户,K个窗口。已知每个客户的到达时间和需要的时长(不超过60min,超过的会被压缩成60min),如果有窗口就依次过去,如果没有窗口就在黄线外等候(黄线外只有一个队伍,先来先服务),求客户的平均等待时长(十进制,单位:分钟)。银行开放时间为8点到17点,再8点之前不开门,8点之前来的人都要等待,在17点后来的人不被服务。
  • < algorithm >中sort()对vector进行排序
  • 读取标准时间用scanf()
  • 精确到秒,先全部转化为秒来计算
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

struct node {
    int come, time;
} customer;

bool cmp1(node a, node b) {
    return a.come < b.come;
}

int main(int argc, const char * argv[]) {
    int N, K;
    cin >> N >> K;
    vector<node> line;
    for(int i = 0; i < N; ++i) {
        int hour, minute, second, time;
        scanf("%d:%d:%d %d", &hour, &minute, &second, &time);
        if(time > 60)
            time = 60;
        int cometime = hour * 3600 + minute * 60 + second;
        if(cometime > (17*60*60)) // 17点
            continue;
        customer.come = cometime;
        customer.time = time * 60;
        line.push_back(customer);
    }
    sort(line.begin(), line.end(), cmp1);
    vector<int> window(K, (8*60*60)); // 存储各窗口的结束时间,大小为K,初始化为8点
    double result = 0.0;
    for(int i = 0; i < line.size(); ++i) {
        int index = 0; // 窗口号
        int minfinish = window[0]; // 各窗口中的最早结束时间

        // 循环检查各窗口,如果有窗口结束时间更早,则更新最早结束时间和窗口号:
        for(int j = 1; j < K; ++j) {
            if(minfinish > window[j]) {
                minfinish = window[j];
                index = j;
            }
        }

        // 确定下一个服务的窗口号和时间后:与客户到达时间进行比较
        if(window[index] <= line[i].come) { // 如果在客户到达之前,客户无等待
            window[index] = line[i].come + line[i].time;
        } else { // 在客户到达之后,更新等待时间result
            result += (window[index] - line[i].come);
            window[index] += line[i].time;
        }
    }
    if(line.size() == 0)
        printf("0.0");
    else
        printf("%.1f", result / 60.0 / line.size());

    return 0;
}

附原题:

Suppose a bank has K 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 T and the processing time P 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: N (<=10000) - the total number of customers, and K (<=100) - the number of windows. Then N lines follow, each contains 2 times: HH:MM:SS - the arriving time, and P - 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
### 银行排队问题的Python实现 银行排队问题是典型的并行处理和任务分配场景,可以通过ZeroMQ框架中的Ventilator-Worker-Sink模型来解决[^1]。以下是基于该模型的一个简单Python实现: #### ZeroMQ Ventilator 实现 Ventilator负责生成任务并将它们发送给Workers。 ```python import zmq import time context = zmq.Context() # Socket to send tasks to workers sender = context.socket(zmq.PUSH) sender.bind("tcp://*:5557") print("Press Enter when the workers are ready...") _ = input() print("Sending tasks to workers...") # Send out tasks total_msec = 0 for task_nbr in range(100): workload = int((task_nbr * task_nbr) % 100 + 1) # Some random work load total_msec += workload sender.send_string(str(workload)) print(f"Total expected cost: {total_msec} msec") time.sleep(1) # Give 0MQ time to deliver ``` #### ZeroMQ Worker 实现 Worker接收来自Ventilator的任务并执行计算后将结果返回给Sink。 ```python import zmq import sys import time context = zmq.Context() # Socket to receive messages on receiver = context.socket(zmq.PULL) receiver.connect("tcp://localhost:5557") # Socket to send messages to sender = context.socket(zmq.PUSH) sender.connect("tcp://localhost:5558") while True: s = receiver.recv_string() print(f"Received request: {s}") # Do some 'work' time.sleep(int(s) / 10) # Send results to sink sender.send(b'') ``` #### ZeroMQ Sink 实现 Sink收集所有Worker的结果,并统计完成时间。 ```python import zmq import time context = zmq.Context() # Socket to collect worker responses receiver = context.socket(zmq.PULL) receiver.bind("tcp://*:5558") # Wait for start of batch s = receiver.recv() # Start our clock now tstart = time.time() # Process 100 confirmations for task_nbr in range(100): s = receiver.recv() if task_nbr % 10 == 0: sys.stdout.write(':') else: sys.stdout.write('.') sys.stdout.flush() # Calculate and report duration of batch tend = time.time() print(f"\nTotal elapsed time: {(tend-tstart)*1000} msec") ``` 上述代码展示了如何通过ZeroMQ构建一个简单的分布式任务管理系统[^2]。对于银行排队问题,可以将其视为多个客户作为任务被分配到不同的柜员(即Worker),而最终的结果由Sink汇总。 此外,在高并发环境下,还需要注意内存管理和I/O性能优化[^3]。建议使用缓存机制减少磁盘操作频率,并利用消息队列实现各模块间的异步通信。
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