生成一段用分支限界法求解八数码问题最优解的代码

好的，以下是使用分支限界法求解八数码问题最优解的 Python 代码：

import heapq

class PuzzleNode:
    def __init__(self, board, moves, priority):
        self.board = board
        self.moves = moves
        self.priority = priority
    
    def __lt__(self, other):
        return self.priority < other.priority
    
    def __eq__(self, other):
        return self.board == other.board
    
    def __hash__(self):
        return hash(str(self.board))
    
    def manhattan_distance(self):
        # 计算曼哈顿距离
        distance = 0
        for i in range(3):
            for j in range(3):
                if self.board[i][j] != 0:
                    row = (self.board[i][j] - 1) // 3
                    col = (self.board[i][j] - 1) % 3
                    distance += abs(i - row) + abs(j - col)
        return distance

class PuzzleSolver:
    def __init__(self, board):
        self.initial_node = PuzzleNode(board, 0, 0)
        self.goal_node = PuzzleNode([[1, 2, 3], [4, 5, 6], [7, 8, 0]], 0, 0)
    
    def solve(self):
        heap = []
        visited = set()
        
        heapq.heappush(heap, self.initial_node)
        
        while heap:
            current_node = heapq.heappop(heap)
            
            if current_node.board == self.goal_node.board:
                return current_node.moves
            
            visited.add(current_node)
            
            moves = current_node.moves + 1
            for next_board in self.get_next_boards(current_node.board):
                next_node = PuzzleNode(next_board, moves, moves + self.get_heuristic(next_board))
                
                if next_node not in visited:
                    heapq.heappush(heap, next_node)
        
        return -1
    
    def get_heuristic(self, board):
        node = PuzzleNode(board, 0, 0)
        return node.manhattan_distance()
    
    def get_next_boards(self, board):
        next_boards = []
        zero_row, zero_col = self.get_zero_position(board)
        
        for direction in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
            next_row, next_col = zero_row + direction[0], zero_col + direction[1]
            
            if 0 <= next_row < 3 and 0 <= next_col < 3:
                next_board = [row[:] for row in board]
                next_board[zero_row][zero_col], next_board[next_row][next_col] = next_board[next_row][next_col], next_board[zero_row][zero_col]
                next_boards.append(next_board)
        
        return next_boards
    
    def get_zero_position(self, board):
        for i in range(3):
            for j in range(3):
                if board[i][j] == 0:
                    return i, j

这个代码使用了优先队列来实现分支限界法，其中 `PuzzleNode` 类表示八数码问题中的一个状态节点，包括当前的棋盘状态、当前的移动步数和优先级。`PuzzleSolver` 类则是求解八数码问题的主要逻辑，包括启发式函数和获取下一个状态的方法。在 `solve` 方法中，我们通过不断扩展优先队列中的节点来搜索最优解。如果队列为空，说明无解，返回 `-1`。如果当前节点是目标节点，返回当前节点的移动步数。否则，我们会生成当前节点的所有下一个状态，计算它们的优先级并加入优先队列中。在加入之前，我们需要确保下一个状态不是已经访问过的状态，以避免重复搜索。