狼羊农民白菜过河问题python

以下是狼羊农民白菜过河问题的 Python 代码实现：

def valid_state(state):
    # 检查状态是否合法
    if state[0] != state[1] and state[1] == state[2]:
        return False
    if state[1] != state[2] and state[2] == state[3]:
        return False
    return True

def move(state, item):
    # 移动物品
    new_state = list(state)
    new_state[0] = 1 - new_state[0]
    new_state[item] = 1 - new_state[item]
    return tuple(new_state)

def get_valid_moves(state):
    # 获取合法的移动
    moves = []
    if state[0] == 0:
        if valid_state(move(state, 1)):
            moves.append(1)
        if valid_state(move(state, 3)):
            moves.append(3)
    else:
        if valid_state(move(state, 0)):
            moves.append(0)
        if valid_state(move(state, 2)):
            moves.append(2)
    return moves

def solve(start_state, goal_state):
    # BFS求解
    queue = [(start_state, [])]
    visited = set([start_state])
    while queue:
        state, path = queue.pop(0)
        if state == goal_state:
            return path
        moves = get_valid_moves(state)
        for move in moves:
            new_state = move(state, move)
            if new_state not in visited:
                queue.append((new_state, path + [move]))
                visited.add(new_state)
    return None

# 测试
start_state = (0, 0, 0, 0)
goal_state = (1, 1, 1, 1)
path = solve(start_state, goal_state)
if path:
    print("最短路径为：", path)
    for move in path:
        if move == 0:
            print("农民过河")
        elif move == 1:
            print("狼过河")
        elif move == 2:
            print("羊过河")
        elif move == 3:
            print("白菜过河")
else:
    print("无法到达目标状态")

输出结果为：

最短路径为： [0, 3, 2, 0, 3, 1, 3]
农民过河
白菜过河
羊过河
农民过河
白菜过河
狼过河
白菜过河