狼羊蔬菜过河python

以下是一个用Python解决狼羊蔬菜过河问题的例子：

w = [1, 2, 3]  # 西岸初始状态，0代表没有，1代表白菜，2代表羊，3代表狼
e = [0] * 3  # 东岸初始状态
li = []  # 储存每一次所带的东西
cnt = 0  # 记录农夫所在位置，单数代表西岸，双数代表东岸

# 条件处理一：农夫不在的一侧，不能同时存在狼和羊，羊和白菜。条件不成立则重新开始。
if cnt % 2 == 1 and (3 in w and 2 in w or 2 in w and 1 in w) or cnt % 2 == 0 and (3 in e and 2 in e or  in e and 1 in e):
    w = [1, 2, 3    e = [0] * 3
    li = []
    cnt = 0

相关问题

人狼羊菜过河 python

人狼羊菜过河是一个经典的智力游戏，玩家需要帮助一群人、一只狼、一只羊和一堆菜过河，但船只只能承载两个物品或人。同时，如果没有人看守，羊会被狼吃掉，而如果没有羊吃菜，菜也会被吃掉。因此，玩家需要想出一个方案，让所有物品和人都安全地渡河。

以下是人狼羊菜过河的 Python 实现示例：

# 定义初始状态
start = {'people': 1, 'wolf': 1, 'sheep': 1, 'vegetable': 1, 'boat': 0}
# 定义目标状态
goal = {'people': 0, 'wolf': 0, 'sheep': 0, 'vegetable': 0, 'boat': 1}

# 定义移动函数
def move(state, p, q):
    new_state = state.copy()
    new_state['people'] -= p
    new_state['wolf'] -= q
    new_state['sheep'] += q
    new_state['vegetable'] += p
    new_state['boat'] = 1 - state['boat']
    return new_state

# 定义判断函数
def is_valid(state):
    if state['wolf'] == state['sheep'] and state['people'] != state['wolf']:
        return False
    if state['sheep'] == state['vegetable'] and state['people'] != state['sheep']:
        return False
    return True

# 定义搜索函数
def search(start, goal):
    visited = set()
    queue = [[start]]
    while queue:
        path = queue.pop(0)
        last_state = path[-1]
        if last_state == goal:
            return path
        for p in range(2):
            for q in range(2):
                if p + q == 0 or p + q == 2:
                    continue
                new_state = move(last_state, p, q)
                if is_valid(new_state) and tuple(new_state.items()) not in visited:
                    visited.add(tuple(new_state.items()))
                    new_path = path + [new_state]
                    queue.append(new_path)

# 运行搜索函数并输出结果
result = search(start, goal)
for state in result:
    print(state)

人狼羊菜过河python，要求用0和1表示是否存在的状态，输出结果为展示人狼羊菜过河的过程

"人狼羊菜过河"是一个经典的逻辑谜题，通常用于讲解搜索算法，如深度优先搜索（DFS）或广度优先搜索（BFS）。在这种情况下，我们可以用Python编写一个简单的程序，用二进制位表示四种角色（人、狼、羊、菜）能否安全过河的状态。每个角色都对应二进制的一位，0代表能过，1代表不能过。

以下是简单版的Python代码示例：

def move_rocks(people_wolves_rabbits_cabbages, moves):
    # 初始化状态数组，全为0表示初始状态可以过河
    states = [0b0000] * (1 << len(moves))

    def dfs(state, index):
        if index == len(moves):  # 所有移动都已经处理完
            return state

        rock = moves[index]
        if rock not in ("P", "W", "R", "C"):  # 非法移动，直接返回当前状态
            return state

        # 计算所有可能的结果
        for new_state in (state | (1 << role) for role in rock):
            # 如果新的状态下所有人狼都能过河，则接受这个状态
            if '1' not in str(bin(new_state)):
                states.append(new_state)
                dfs(new_state, index + 1)

    dfs(people_wolves_rabbits_cabbages, 0)
    return states

# 示例输入：人（P）、狼（W）、羊（R）、菜（C），以及一组移动记录（例如 ["PC", "RP", ...]）
people_wolves_rabbits_cabbages = "PWRC"
moves = ["PC", "RP"]
result = move_rocks(int("".join(people_wolves_rabbits_cabbages), 2), 0)
for i, state in enumerate(result, start=1):
    print(f"步骤 {i}: {''.join(['人狼羊菜'[bit] for bit in bin(state)[2:].zfill(4)])}")