农夫过河问题python

好的，农夫过河问题是一个经典的逻辑问题，题目描述如下：

一个农夫要带着一只狼、一只羊和一颗白菜过河，但是小船只能装下农夫和另外一件物品。如果农夫不在场，狼就会吃羊，羊不在场，白菜就会被吃。请问农夫如何才能安全地将这三样东西全部运到河的对岸？

以下是一个简单的 Python 实现：

class State:
    def __init__(self, farmer, wolf, goat, cabbage, boat):
        self.farmer = farmer  # 0:在左岸，1:在右岸
        self.wolf = wolf
        self.goat = goat
        self.cabbage = cabbage
        self.boat = boat  # 0:在左岸，1:在右岸

    def is_valid(self):
        if self.farmer not in [0, 1] or self.wolf not in [0, 1] or self.goat not in [0, 1] or self.cabbage not in [0, 1] or self.boat not in [0, 1]:
            return False
        if self.wolf == self.goat and self.farmer != self.wolf:
            return False
        if self.goat == self.cabbage and self.farmer != self.goat:
            return False
        return True

    def is_goal(self):
        return self.farmer == 1 and self.wolf == 1 and self.goat == 1 and self.cabbage == 1

    def __eq__(self, other):
        return self.farmer == other.farmer and self.wolf == other.wolf and self.goat == other.goat and self.cabbage == other.cabbage and self.boat == other.boat

    def __hash__(self):
        return hash((self.farmer, self.wolf, self.goat, self.cabbage, self.boat))


class Node:
    def __init__(self, state, parent=None, action=None):
        self.state = state
        self.parent = parent
        self.action = action

    def expand(self):
        children = []
        for a in ["farmer", "wolf", "goat", "cabbage"]:
            new_state = State(self.state.farmer, self.state.wolf, self.state.goat, self.state.cabbage, self.state.boat)
            setattr(new_state, a, 1 - getattr(self.state, a))  # 改变属性值
            new_state.boat = 1 - self.state.boat  # 船也过河了
            child = Node(new_state, self, a)
            if new_state.is_valid():
                children.append(child)
        return children


def bfs(initial_state):
    initial_node = Node(initial_state)
    if initial_state.is_goal():
        return initial_node

    frontier = [initial_node]
    explored = set()

    while frontier:
        node = frontier.pop(0)
        explored.add(node.state)
        for child in node.expand():
            if child.state.is_goal():
                return child
            if child.state not in explored and child not in frontier:
                frontier.append(child)

    return None


initial_state = State(0, 0, 0, 0, 0)
solution = bfs(initial_state)

if not solution:
    print("No solution found!")
else:
    path = []
    node = solution
    while node:
        path.append(node)
        node = node.parent
    path.reverse()
    for node in path:
        print(node.state.farmer, node.state.wolf, node.state.goat, node.state.cabbage, node.state.boat, node.action)

这里使用了 BFS 算法来搜索解决方案，首先定义了 `State` 类来表示状态，包含农夫、狼、羊、白菜和船的位置，以及一些辅助方法。然后定义了 `Node` 类来表示搜索树中的节点，包含一个状态和一个指向父节点的指针。最后定义了 `bfs` 函数来执行 BFS 算法搜索解决方案，返回最终状态的节点。