多个电梯代码

以下是一个简单的多电梯调度的Python代码实现：

class Elevator:
    def __init__(self, id, floor=1, direction=1, status='idle'):
        self.id = id
        self.floor = floor
        self.direction = direction   # 1:up, -1:down
        self.status = status         # idle, up, down, stop
        self.requests = []           # up/down floor requests
        self.passengers = []         # passengers in elevator
        self.capacity = 10           # max capacity
        self.stops = []              # stops history

    def __str__(self):
        return f"Elevator {self.id} at floor {self.floor} ({self.status})"

    def add_request(self, floor, direction):
        if floor == self.floor:
            return
        if self.status == 'idle':
            self.status = 'up' if floor > self.floor else 'down'
        elif self.status == 'up' and direction == -1:
            self.requests.append(floor)
            return
        elif self.status == 'down' and direction == 1:
            self.requests.append(floor)
            return
        elif direction == self.direction:
            if (self.status == 'up' and floor > self.floor) or (self.status == 'down' and floor < self.floor):
                self.requests.append(floor)
                return
        self.requests.append(floor)
        self.direction = 1 if floor > self.floor else -1

    def move(self):
        if self.status == 'up':
            self.floor += 1
        elif self.status == 'down':
            self.floor -= 1
        self.stops.append(self.floor)
        if self.floor in self.requests:
            self.requests.remove(self.floor)
            self.passengers += [p for p in self.floor_passengers() if p not in self.passengers]
            if len(self.passengers) >= self.capacity or len(self.requests) == 0:
                self.status = 'stop'
        if len(self.requests) == 0 and self.status != 'idle':
            if self.floor not in self.stops:
                self.stops.append(self.floor)
            self.status = 'idle'

    def floor_passengers(self):
        return [p for p in self.requests if (p - self.floor) * self.direction > 0]

class ElevatorController:
    def __init__(self, num_elevators=1, num_floors=5):
        self.elevators = [Elevator(i+1) for i in range(num_elevators)]
        self.num_floors = num_floors

    def __str__(self):
        return '\n'.join([str(e) for e in self.elevators])

    def status(self):
        return '\n'.join([f"Elevator {e.id} at floor {e.floor} ({e.status}) with {len(e.passengers)} passengers" for e in self.elevators])

    def add_request(self, floor, direction):
        elevators = sorted(self.elevators, key=lambda x: abs(x.floor - floor))
        for e in elevators:
            e.add_request(floor, direction)

    def step(self):
        for e in self.elevators:
            e.move()

if __name__ == '__main__':
    controller = ElevatorController(num_elevators=2, num_floors=10)

    # add requests
    controller.add_request(3, 1)
    controller.add_request(5, -1)
    controller.add_request(7, 1)

    # run simulation
    for i in range(20):
        print(f"Step {i}")
        print(controller.status())
        controller.step()

这段代码实现了一个ElevatorController类，其中包含多个Elevator对象，可以通过add_request()方法来添加上下行请求，通过step()方法来模拟电梯运行过程，最后输出每个电梯的状态。具体实现中，每个电梯会根据当前状态和乘客请求来判断下一步的行动，同时记录每个电梯的停靠历史和乘客信息。