将多个不同长、宽、高、重量的集装箱装入两个不同长、宽、高、载重限制的货船，并画出装载图。PYTHON

这是一个 NP-完全问题，也就是说没有多项式时间的解法。我们可以使用贪心算法来近似解决这个问题。以下是 Python 代码：

class Container:
    def __init__(self, length, width, height, weight):
        self.length = length
        self.width = width
        self.height = height
        self.weight = weight

class Ship:
    def __init__(self, length, width, height, weight_limit):
        self.length = length
        self.width = width
        self.height = height
        self.weight_limit = weight_limit
        self.containers = []

    def load_container(self, container):
        if (len(self.containers) == 0):
            self.containers.append([container])
            return True

        for stack in self.containers:
            if (self.can_load(stack, container)):
                stack.append(container)
                return True

        if (self.can_load_new_stack(container)):
            self.containers.append([container])
            return True

        return False

    def can_load(self, stack, container):
        if (sum(c.weight for c in stack) + container.weight > self.weight_limit):
            return False

        length = max(c.length for c in stack)
        width = max(c.width for c in stack)
        height = sum(c.height for c in stack) + container.height
        if (length > self.length or width > self.width or height > self.height):
            return False

        return True

    def can_load_new_stack(self, container):
        if (container.weight > self.weight_limit or
            container.length > self.length or
            container.width > self.width or
            container.height > self.height):
            return False

        return True

def load_containers(containers, ships):
    ships.sort(key=lambda s: s.weight_limit, reverse=True)

    for container in containers:
        loaded = False
        for ship in ships:
            if (ship.load_container(container)):
                loaded = True
                break

        if (not loaded):
            return False

    return True

# Example usage
c1 = Container(10, 5, 3, 2)
c2 = Container(8, 4, 4, 3)
c3 = Container(6, 3, 2, 1)
c4 = Container(7, 4, 5, 2)
c5 = Container(9, 6, 4, 3)

s1 = Ship(20, 8, 6, 10)
s2 = Ship(15, 6, 5, 7)

containers = [c1, c2, c3, c4, c5]
ships = [s1, s2]

if (load_containers(containers, ships)):
    for i, ship in enumerate(ships):
        print("Ship", i+1)
        for j, stack in enumerate(ship.containers):
            print("Stack", j+1)
            for container in stack:
                print(container.length, container.width, container.height, container.weight)
else:
    print("Not enough space to load all containers")

这个算法的基本思路是，对于每个集装箱，我们尝试将其装入能够承载它的载重限制最大的船只中。如果没有可以承载它的船只，就返回 False。我们使用一个二维列表来存储每个船只中的集装箱，其中每个子列表表示一个货物堆。最后，我们输出每个船只中的所有货物堆和其中的集装箱。