python 开发一个 A星算法接口

好的，下面是一个简单的 Python 实现 A* 算法的代码，仅供参考：

from queue import PriorityQueue

# 状态类
class State:
    def __init__(self, board, moves, parent=None, cost=0):
        self.board = board
        self.moves = moves
        self.parent = parent
        self.cost = cost
        
    # 计算启发函数
    def heuristic(self, goal):
        count = 0
        for i in range(3):
            for j in range(3):
                if self.board[i][j] != goal[i][j]:
                    count += 1
        return count
        
    # 计算总代价
    def total_cost(self, goal):
        return self.cost + self.heuristic(goal)
    
    # 扩展状态
    def expand(self, goal):
        children = []
        for i in range(3):
            for j in range(3):
                if self.board[i][j] == 0:
                    if i > 0:
                        new_board = self.move(i, j, i - 1, j)
                        children.append(State(new_board, self.moves + 'U', self, self.cost + 1).total_cost(goal))
                    if i < 2:
                        new_board = self.move(i, j, i + 1, j)
                        children.append(State(new_board, self.moves + 'D', self, self.cost + 1).total_cost(goal))
                    if j > 0:
                        new_board = self.move(i, j, i, j - 1)
                        children.append(State(new_board, self.moves + 'L', self, self.cost + 1).total_cost(goal))
                    if j < 2:
                        new_board = self.move(i, j, i, j + 1)
                        children.append(State(new_board, self.moves + 'R', self, self.cost + 1).total_cost(goal))
        return children
    
    # 移动
    def move(self, i1, j1, i2, j2):
        new_board = [[self.board[i][j] for j in range(3)] for i in range(3)]
        new_board[i1][j1], new_board[i2][j2] = new_board[i2][j2], new_board[i1][j1]
        return new_board

# A* 算法求解
def solve(initial, goal):
    initial_state = State(initial, '', None, 0)
    goal_state = State(goal, None, None)
    visited = set()
    pq = PriorityQueue()
    pq.put((initial_state.total_cost(goal), initial_state))
    while not pq.empty():
        _, state = pq.get()
        if state.board == goal:
            return state.moves
        children = state.expand(goal)
        for child in children:
            if str(child.board) not in visited:
                pq.put((child, child.total_cost(goal)))
                visited.add(str(child.board))

# 测试
if __name__ == '__main__':
    initial = [[1, 2, 3], [4, 0, 6], [7, 5, 8]]
    goal = [[1, 2, 3], [4, 5, 6], [7, 8, 0]]
    moves = solve(initial, goal)
    print(moves)

上面的代码中，我们定义了一个状态类 `State`，封装了状态的信息以及对状态进行操作的方法。然后，我们实现了 A* 算法的求解函数 `solve`。最后，在 `main` 函数中，我们构造了一个初始状态和目标状态，并用求解函数求解出了步骤。