我需要基于A星算法求解多机器人栅格路径规划的python代码

好的，以下是一个简单的基于A星算法的多机器人栅格路径规划Python代码示例：

import heapq

class Robot:
    def __init__(self, start, goal):
        self.start = start
        self.goal = goal
        
class Grid:
    def __init__(self, width, height, obstacles, robots):
        self.width = width
        self.height = height
        self.obstacles = obstacles
        self.robots = robots
        
    def is_obstacle(self, x, y):
        return (x, y) in self.obstacles
        
    def is_goal(self, x, y):
        for robot in self.robots:
            if robot.goal == (x, y):
                return True
        return False
        
    def cost(self, x1, y1, x2, y2):
        return 1
        
class Node:
    def __init__(self, x, y, g, h, parent):
        self.x = x
        self.y = y
        self.g = g
        self.h = h
        self.parent = parent
        
    def f(self):
        return self.g + self.h

def a_star(grid, robots):
    start_nodes = [Node(robot.start[0], robot.start[1], 0, 0, None) for robot in robots]
    open_set = []
    closed_set = set()
    for start_node in start_nodes:
        heapq.heappush(open_set, (start_node.f(), start_node))
    while open_set:
        _, current_node = heapq.heappop(open_set)
        if all(grid.is_goal(current_node.x, current_node.y) for robot in robots):
            paths = []
            for i, robot in enumerate(robots):
                path = []
                node = current_node
                while node:
                    path.append((node.x, node.y))
                    node = node.parent[i]
                paths.append(path[::-1])
            return paths
        closed_set.add((current_node.x, current_node.y))
        for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
            x = current_node.x + dx
            y = current_node.y + dy
            g = current_node.g + grid.cost(current_node.x, current_node.y, x, y)
            h = sum(manhattan_distance(x, y, robot.goal[0], robot.goal[1]) for robot in robots)
            node = Node(x, y, g, h, current_node)
            if x < 0 or x >= grid.width or y < 0 or y >= grid.height \
                    or grid.is_obstacle(x, y) or (x, y) in closed_set:
                continue
            heapq.heappush(open_set, (node.f(), node))
    return None

def manhattan_distance(x1, y1, x2, y2):
    return abs(x1 - x2) + abs(y1 - y2)

# 示例用法
obstacles = [(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)]
robots = [Robot((1, 1), (6, 5)), Robot((1, 3), (6, 1))]
grid = Grid(7, 6, obstacles, robots)
paths = a_star(grid, robots)
print(paths)

其中，`Robot` 类表示一个机器人，包含起点和终点坐标；`Grid` 类表示栅格地图，包含宽度、高度、障碍物坐标和机器人列表；`Node` 类表示 A* 算法中的节点，包含坐标、从起点到该点的代价、估价函数值和父节点。`a_star` 函数使用 A* 算法求解多机器人栅格路径规划问题，返回每个机器人的路径。`manhattan_distance` 函数计算两点之间的曼哈顿距离。在示例中，定义了一个 7x6 的栅格地图，其中包含了一些障碍物和两个机器人，调用 `a_star` 函数求解路径并输出结果。