三个商人和三个仆人过河代码

这是一个经典的过河问题，以下是三个商人和三个仆人过河的代码实现：

def cross_river(state, path):
    if state == ([], [], [], []):
        return path
    for move in moves:
        if is_valid(state, move):
            new_state = update_state(state, move)
            if new_state not in visited:
                visited.add(new_state)
                new_path = path + [move]
                result = cross_river(new_state, new_path)
                if result:
                    return result
    return None

def is_valid(state, move):
    left_bank, right_bank, left_boat, right_boat = state
    l1, l2, r1, r2 = move
    if l1 in right_bank or l2 in right_bank or r1 in left_bank or r2 in left_bank:
        return False
    if len(left_boat) + len([l1, l2]) > 2 or len(right_boat) + len([r1, r2]) > 2:
        return False
    if len(left_boat) == 0 and (r1 in right_bank or r2 in right_bank):
        return False
    if len(right_boat) == 0 and (l1 in left_bank or l2 in left_bank):
        return False
    return True

def update_state(state, move):
    left_bank, right_bank, left_boat, right_boat = state
    l1, l2, r1, r2 = move
    if len(left_boat) == 0:
        left_boat += [l1, l2]
        left_bank.remove(l1)
        left_bank.remove(l2)
    else:
        left_boat += [r1, r2]
        left_bank.remove(r1)
        left_bank.remove(r2)
    if len(right_boat) == 0:
        right_boat += [r1, r2]
        right_bank.remove(r1)
        right_bank.remove(r2)
    else:
        right_boat += [l1, l2]
        right_bank.remove(l1)
        right_bank.remove(l2)
    return (left_bank, right_bank, left_boat, right_boat)

moves = [(x, y, z, w) for x in range(3) for y in range(3) for z in range(3) for w in range(3) if x+y+z+w == 2 and len(set([x, y, z, w])) == 2]
visited = set()
state = ([0, 0, 0], [1, 1, 1], [], [])
path = []
result = cross_river(state, path)
print(result)

其中，`state`表示当前状态，`(left_bank, right_bank, left_boat, right_boat)`分别表示左岸、右岸、左船和右船上的人员情况。`moves`表示所有可能的移动方式，`visited`表示已经访问过的状态，`path`表示当前路径。`is_valid`函数用于判断当前移动是否合法，`update_state`函数用于更新状态。最后，使用递归函数`cross_river`求解最短路径。