用python完成A*算法

好的，以下是Python实现A*算法的示例代码：

# 定义节点类
class Node:
    def __init__(self, parent=None, position=None):
        self.parent = parent
        self.position = position

        self.g = 0  # 从起点到当前节点的实际代价
        self.h = 0  # 从当前节点到目标节点的估计代价
        self.f = 0  # f = g + h

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


# 定义A*算法函数
def astar(maze, start, end):
    # 初始化起点和终点
    start_node = Node(None, start)
    start_node.g = start_node.h = start_node.f = 0
    end_node = Node(None, end)
    end_node.g = end_node.h = end_node.f = 0

    # 初始化开启列表和关闭列表
    open_list = []
    closed_list = []

    # 将起点加入开启列表
    open_list.append(start_node)

    # 循环直到找到终点或者开启列表为空
    while len(open_list) > 0:
        # 获取f值最小的节点
        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

        # 将当前节点从开启列表中移除，并加入关闭列表
        open_list.pop(current_index)
        closed_list.append(current_node)

        # 判断是否找到终点
        if current_node == end_node:
            path = []
            current = current_node
            while current is not None:
                path.append(current.position)
                current = current.parent
            return path[::-1]  # 返回路径，倒序

        # 获取当前节点的所有邻居
        neighbors = []
        for new_position in [(0, -1), (0, 1), (-1, 0), (1, 0), (-1, -1), (-1, 1), (1, -1), (1, 1)]:
            node_position = (current_node.position[0] + new_position[0], current_node.position[1] + new_position[1])

            # 确保节点在迷宫内
            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

            # 确保节点不是墙
            if maze[node_position[0]][node_position[1]] != 0:
                continue

            # 创建新节点
            new_node = Node(current_node, node_position)

            # 添加邻居
            neighbors.append(new_node)

        # 遍历所有邻居
        for neighbor in neighbors:
            # 如果邻居已经在关闭列表中，跳过
            if neighbor in closed_list:
                continue

            # 计算邻居的f、g和h值
            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

            # 如果邻居已经在开启列表中，并且新的g值更大，跳过
            for open_node in open_list:
                if neighbor == open_node and neighbor.g > open_node.g:
                    continue

            # 将邻居加入开启列表
            open_list.append(neighbor)

其中maze表示迷宫，0表示可走，其他数字表示墙；start和end分别表示起点和终点的坐标。调用该函数即可得到从起点到终点的最短路径。