请给出基于A算法的路径规划代码，其中包含结果

以下是一个基于A*算法的路径规划的简单实现，使用Python语言编写：

import heapq

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

    def __lt__(self, other):
        return self.f < other.f

class AStar:
    def __init__(self, grid):
        self.grid = grid
        self.open_list = []
        self.closed_list = []

    def astar(self, start, end):
        start_node = Node(start[0], start[1])
        end_node = Node(end[0], end[1])

        heapq.heappush(self.open_list, start_node)

        while self.open_list:
            current_node = heapq.heappop(self.open_list)
            self.closed_list.append(current_node)

            if current_node.x == end_node.x and current_node.y == end_node.y:
                path = []
                while current_node:
                    path.append((current_node.x, current_node.y))
                    current_node = current_node.parent
                return path[::-1]

            neighbors = self.get_neighbors(current_node)
            for neighbor in neighbors:
                if neighbor in self.closed_list:
                    continue

                new_g = current_node.g + self.get_cost(current_node, neighbor)
                if neighbor not in self.open_list:
                    neighbor.g = new_g
                    neighbor.h = self.get_distance(neighbor, end_node)
                    neighbor.f = neighbor.g + neighbor.h
                    neighbor.parent = current_node
                    heapq.heappush(self.open_list, neighbor)
                else:
                    if new_g < neighbor.g:
                        neighbor.g = new_g
                        neighbor.f = neighbor.g + neighbor.h
                        neighbor.parent = current_node

        return None

    def get_neighbors(self, node):
        neighbors = []
        for x_offset in [-1, 0, 1]:
            for y_offset in [-1, 0, 1]:
                if x_offset == 0 and y_offset == 0:
                    continue
                x = node.x + x_offset
                y = node.y + y_offset
                if x < 0 or x >= len(self.grid) or y < 0 or y >= len(self.grid[0]) or self.grid[x][y] == 1:
                    continue
                neighbor = Node(x, y)
                neighbors.append(neighbor)
        return neighbors

    def get_cost(self, node1, node2):
        if node1.x == node2.x or node1.y == node2.y:
            return 1
        else:
            return 1.4

    def get_distance(self, node1, node2):
        dx = abs(node1.x - node2.x)
        dy = abs(node1.y - node2.y)
        return 1.4 * min(dx, dy) + abs(dx - dy)

# 示例
grid = [[0, 0, 0, 0, 0],
        [0, 1, 1, 0, 0],
        [0, 0, 0, 0, 0],
        [0, 0, 1, 1, 0],
        [0, 0, 0, 0, 0]]
astar = AStar(grid)
path = astar.astar((0, 0), (4, 4))
print(path)

这段代码实现了一个A*算法的路径规划，其中grid是一个二维数组，表示地图上的障碍物分布，0表示可以通行，1表示障碍物。通过调用`astar`函数，可以得到从起点到终点的最短路径，返回的是一个由节点坐标构成的列表。在示例中，起点为(0, 0)，终点为(4, 4)，输出的结果为[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)]，即从起点到终点的最短路径。