1017 Queueing at Bank

struct Customer
{
    int arrive_time, cost_time, start_time, end_time, wait_time = 0;
};
int N, K;

bool cmp(const Customer &c1, const Customer &c2)
{
    return c1.arrive_time < c2.arrive_time;
}

int count_seconds(string st)
{
    int h = stoi(st.substr(0, 2));
    int m = stoi(st.substr(3, 2));
    int s = stoi(st.substr(6, 2));
    return h * 60 * 60 + m * 60 + s;
}

int main()
{
    cin >> N >> K;
    int open_time = 8 * 60 * 60, close_time = 17 * 60 * 60;

    // 读取所有顾客,以便按到达时间排序
    vector<Customer> customers(N);
    for(int i = 0; i < N; ++i)
    {
        string come_time;
        int p_time;
        cin >> come_time >> p_time;
        Customer customer;
        customer.arrive_time = count_seconds(come_time);
        customer.cost_time = p_time * 60;

        customers[i] = customer;
    }

    // 按到达时间排序
    sort(customers.begin(), customers.end(), cmp);
    double sum_wait = 0;

    int valid = N;

    // k个窗口,k个队列
    vector<queue<Customer>> windows(K);
    for(int i = 0; i < N; ++i)
    {
        Customer &cur_user = customers[i];

        // 17:00之后到达的顾客不计算
        if(cur_user.arrive_time > close_time)
        {
            valid = i;
            break;
        }

        // 先排空窗口
        if(i < K)
        {
        	// 是否再8:00之前到达
            if(cur_user.arrive_time < open_time)
                cur_user.start_time = open_time;
            else
                cur_user.start_time = cur_user.arrive_time;

            cur_user.end_time = cur_user.start_time + cur_user.cost_time;
            cur_user.wait_time = cur_user.start_time - cur_user.arrive_time;
            
            windows[i].push(cur_user);
            sum_wait += cur_user.wait_time;
        }
        else    // 再找最早的可用窗口
        {
            int win = 0;
            int last_end_time = windows[win].front().end_time;
            for(int i = 1; i < K; ++i)
            {
                if(windows[i].front().end_time < last_end_time)
                {
                    win = i;
                    last_end_time = windows[win].front().end_time;
                }
            }
            windows[win].pop();

            // 计算各时间
            if(cur_user.arrive_time > last_end_time)
                cur_user.start_time = cur_user.arrive_time;
            else
                cur_user.start_time = last_end_time;

            cur_user.end_time = cur_user.start_time + cur_user.cost_time;
            cur_user.wait_time = cur_user.start_time - cur_user.arrive_time;

            windows[win].push(cur_user);
            sum_wait += cur_user.wait_time;
        }
    }
    printf("%.1f", sum_wait / valid / 60.0);
    return 0;
}
### 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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