用Python写一个用a*算法解决八数码问题

以下是用Python实现A*算法解决八数码问题的代码：

import heapq

class State:
    def __init__(self, board, goal, moves=0):
        self.board = board
        self.goal = goal
        self.moves = moves

    def get_new_board(self, i1, i2, moves):
        new_board = self.board[:]
        new_board[i1], new_board[i2] = new_board[i2], new_board[i1]
        return State(new_board, self.goal, moves)

    def get_moves(self, zero_pos):
        moves = []
        if zero_pos > 2:
            moves.append(self.get_new_board(zero_pos, zero_pos-3, self.moves+1))
        if zero_pos < 6:
            moves.append(self.get_new_board(zero_pos, zero_pos+3, self.moves+1))
        if zero_pos % 3 > 0:
            moves.append(self.get_new_board(zero_pos, zero_pos-1, self.moves+1))
        if zero_pos % 3 < 2:
            moves.append(self.get_new_board(zero_pos, zero_pos+1, self.moves+1))
        return moves

    def heuristic(self):
        return sum(abs(b%3 - g%3) + abs(b//3 - g//3)
                   for b, g in ((self.board.index(i), self.goal.index(i)) for i in range(1, 9)))

    def __lt__(self, other):
        return self.heuristic() + self.moves < other.heuristic() + other.moves

    def __eq__(self, other):
        return self.board == other.board

def solve(initial_state):
    heap = []
    heapq.heappush(heap, initial_state)
    closed = set()
    moves = {(str(initial_state.board))}
    count = 0
    while heap:
        state = heapq.heappop(heap)
        if state.board == state.goal:
            return state.moves
        closed.add(str(state.board))
        count += 1
        for move in state.get_moves(state.board.index(0)):
            if str(move.board) not in closed and str(move.board) not in moves:
                moves.add(str(move.board))
                heapq.heappush(heap, move)
    return None

board = [1, 2, 3, 4, 0, 5, 6, 7, 8]
goal = [1, 2, 3, 4, 5, 6, 7, 8, 0]
initial_state = State(board, goal)
print(solve(initial_state))

在这段代码中，我们首先定义了一个State类来表示八数码问题的状态。这个类包括一个代表当前状态的board，一个代表目标状态的goal，以及一个代表移动步数的moves。我们还定义了一些辅助方法来获取新的状态、移动、评估函数等。

在solve函数中，我们使用一个优先队列来存储待扩展的状态。每次从队列中选取评估函数值最小的状态进行扩展，直到找到目标状态。在搜索过程中，我们使用closed集合来存储已经扩展过的状态，避免重复扩展。我们还使用moves集合来存储已经扩展过的状态的字符串表示，同样是为了避免重复扩展。

最后，我们定义了一个初始状态和目标状态，并调用solve函数来解决八数码问题。执行完成后，输出移动步数即可。