迷宫问题的A*算法(python实现)

好的，这个问题可以回答。以下是一份简单的Python实现：

class Node:
    def __init__(self, position, parent=None):
        self.position = position
        self.g = 0
        self.h = 0
        self.f = 0
        self.parent = parent

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

def astar(maze, start, end):
    start_node = Node(start)
    end_node = Node(end)

    open_list = [start_node]
    closed_list = []

    while open_list:
        current_node = open_list[0]
        current_index = 0
        for index, item in enumerate(open_list):
            if item.f < current_node.f:
                current_node = item
                current_index = index

        open_list.pop(current_index)
        closed_list.append(current_node)

        if current_node == end_node:
            path = []
            current = current_node
            while current is not None:
                path.append(current.position)
                current = current.parent
            return path[::-1]

        neighbors = []
        for new_position in [(0, -1), (0, 1), (-1, 0), (1, 0)]:
            node_position = (current_node.position[0] + new_position[0], current_node.position[1] + new_position[1])

            if node_position[0] > (len(maze) - 1) or node_position[0] < 0 or node_position[1] > (len(maze[len(maze)-1]) -1) or node_position[1] < 0:
                continue

            if maze[node_position[0]][node_position[1]] != 0:
                continue

            new_node = Node(node_position, current_node)

            neighbors.append(new_node)

        for neighbor in neighbors:
            if neighbor in closed_list:
                continue

            neighbor.g = current_node.g + 1
            neighbor.h = ((neighbor.position[0] - end_node.position[0]) ** 2) + ((neighbor.position[1] - end_node.position[1]) ** 2)
            neighbor.f = neighbor.g + neighbor.h

            if neighbor in open_list:
                if neighbor.g > item.g:
                    continue

            open_list.append(neighbor)

    return None

它实现了A*算法来解决迷宫问题。你可以将迷宫表示为2D列表，其中每个元素可以是0（表示该位置可以通过）或1（表示该位置是障碍），然后调用此函数来计算从起点到终点的最短路径。