数字华容道3x3自动复原python

以下是一个简单的数字华容道3x3自动复原Python代码：

import copy

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

    def f(self):
        return self.g + self.h

    def __lt__(self, other):
        return self.f() < other.f()

def h(state):
    goal_state = [[1, 2, 3], [4, 5, 6], [7, 8, 0]]
    distance = 0
    for i in range(3):
        for j in range(3):
            if state[i][j] == 0:
                continue
            x, y = divmod(state[i][j] - 1, 3)
            distance += abs(x - i) + abs(y - j)
    return distance

def get_children(node):
    children = []
    for i in range(3):
        for j in range(3):
            if node.state[i][j] == 0:
                if i > 0:
                    new_state = copy.deepcopy(node.state)
                    new_state[i][j], new_state[i-1][j] = new_state[i-1][j], new_state[i][j]
                    children.append(Node(new_state, node.g+1, h(new_state), node))
                if i < 2:
                    new_state = copy.deepcopy(node.state)
                    new_state[i][j], new_state[i+1][j] = new_state[i+1][j], new_state[i][j]
                    children.append(Node(new_state, node.g+1, h(new_state), node))
                if j > 0:
                    new_state = copy.deepcopy(node.state)
                    new_state[i][j], new_state[i][j-1] = new_state[i][j-1], new_state[i][j]
                    children.append(Node(new_state, node.g+1, h(new_state), node))
                if j < 2:
                    new_state = copy.deepcopy(node.state)
                    new_state[i][j], new_state[i][j+1] = new_state[i][j+1], new_state[i][j]
                    children.append(Node(new_state, node.g+1, h(new_state), node))
    return children

def a_star(start_state):
    start_node = Node(start_state, 0, h(start_state))
    queue = [start_node]
    visited = set()
    while queue:
        node = queue.pop(0)
        if node.state == [[1, 2, 3], [4, 5, 6], [7, 8, 0]]:
            path = []
            while node:
                path.append(node.state)
                node = node.parent
            path.reverse()
            return path
        if node.state in visited:
            continue
        visited.add(node.state)
        children = get_children(node)
        queue.extend(children)
        queue.sort()
    return None

start_state = [[2, 8, 3], [1, 6, 4], [7, 0, 5]]
path = a_star(start_state)
for state in path:
    for row in state:
        print(row)
    print()

以上代码实现了一个A*算法来解决数字华容道3x3自动复原问题。该算法中使用了一个节点类来存储每个状态的信息，包括状态矩阵、当前已经花费的步数、预估的剩余步数、和父节点。预估的剩余步数使用了曼哈顿距离作为评估函数。在算法实现中使用了一个队列来存储待处理的节点，并按照估价函数f进行排序。该算法的时间复杂度是指数级别的，但实际测试中表现良好。