我希望有多个线程使用相同的资源。
我正在研究 Single lane bridge problem 的变体，其中最多三辆车可以同时过桥，只要它们按照进入桥的顺序离开桥，并且只要所有车辆都通过朝着同一个方向行驶。
在我的版本中，我希望最多 n 个线程可以访问公共资源。
而且我希望这些线程中的每一个都以类似 FIFO 的方式以正确的顺序完成计算。
有没有办法添加一个大小为 Queue 的 n 来代表桥梁，
然后加起来 n 个踏板？
到目前为止，这是我的工作代码。
现在的样子，左边的车先过桥，然后右边的。
from threading import Lock, Thread, Semaphore
import logging
import time
class ThreadsProcessing:
def __init__(self):
# Threads list.
self.threads = []
# Lock objects.
self.lock_enter_bridge = Lock()
self.lock_exit_bridge = Lock()
self.bridge_sem = Semaphore(0)
def left_side_car(self):
print('Car from left side is starting ...')
print('Car from left is waiting to acquire permission to enter bridge.')
self.lock_enter_bridge.acquire()
print('Car from left has acquired permission to enter bridge, crossing bridge...')
time.sleep(2)
print('Car from left is waiting to acquire lock to exit bridge.')
self.lock_exit_bridge.acquire()
print('Car from left has acquired lock to exit bridge')
print('Car from left has exited the bridge and will release both locks.')
self.lock_enter_bridge.release()
self.lock_exit_bridge.release()
print("Car from left tells the semaphore to the bridge semaphore that its finished with crossing the bridge")
self.bridge_sem.release()
def right_side_car(self):
logging.info('Car from right side is starting ...')
logging.info('Car from right is waiting to acquire permission to enter bridge.')
# Two workers, wait for both of them
numOfWorkers = 2
for _ in range(numOfWorkers):
print('Waiting for bridge semaphore...')
self.bridge_sem.acquire()
print("Car from right got the bridge semaphore")
logging.info("Car from right is crossing the bridge...")
time.sleep(4)
print("Car from right has finished crossing the bridge all done")
self.bridge_sem.release()
def start_car_threads(self):
for thread_func in [self.left_side_car, self.right_side_car]:
self.threads.append(Thread(target=thread_func))
self.threads[-1].start()
def join_threads(self):
for thread in self.threads:
thread.join()
if __name__ == '__main__':
car_threads = ThreadsProcessing()
car_threads.start_car_threads()
car_threads.join_threads()
logging.info('Finished')
问题的完整描述是这样的： 单车道桥最多可支撑三辆朝同一方向行驶的汽车。提供无饥饿的实现。
i) 每辆车开始向桥的左边缘（L）或右边缘（R）移动，这取决于汽车的方向值
ii)每辆车需要一个随机时间 t_arrive 过桥。
iii)当汽车到达桥的左（L）或右（R）边缘时，必要的过程将使用必要的参数调用方法到达桥
