生成用于ros中路径搜索的A*算法

以下是一个简单的A*算法实现，用于在ROS中进行路径搜索：

import rospy
from nav_msgs.msg import OccupancyGrid
from geometry_msgs.msg import PoseStamped
from heapq import heappush, heappop

class AStarPlanner:
    def __init__(self):
        self.map = None
        self.start = None
        self.goal = None

    def get_neighbors(self, x, y):
        neighbors = []
        for dx in [-1, 0, 1]:
            for dy in [-1, 0, 1]:
                if dx == 0 and dy == 0:
                    continue
                nx, ny = x + dx, y + dy
                if nx < 0 or nx >= self.map.info.width or ny < 0 or ny >= self.map.info.height:
                    continue
                if self.map.data[ny * self.map.info.width + nx] != 0:
                    continue
                neighbors.append((nx, ny))
        return neighbors

    def get_cost(self, x1, y1, x2, y2):
        if x1 == x2 or y1 == y2:
            return 1.0
        else:
            return 1.4

    def heuristic(self, x, y):
        return abs(x - self.goal.x) + abs(y - self.goal.y)

    def plan(self):
        if self.map is None or self.start is None or self.goal is None:
            return None

        frontier = [(0.0, self.start.x, self.start.y)]
        came_from = {}
        cost_so_far = {}
        came_from[(self.start.x, self.start.y)] = None
        cost_so_far[(self.start.x, self.start.y)] = 0.0

        while frontier:
            _, x, y = heappop(frontier)
            if x == self.goal.x and y == self.goal.y:
                break

            for nx, ny in self.get_neighbors(x, y):
                new_cost = cost_so_far[(x, y)] + self.get_cost(x, y, nx, ny)
                if (nx, ny) not in cost_so_far or new_cost < cost_so_far[(nx, ny)]:
                    cost_so_far[(nx, ny)] = new_cost
                    priority = new_cost + self.heuristic(nx, ny)
                    heappush(frontier, (priority, nx, ny))
                    came_from[(nx, ny)] = (x, y)

        if (self.goal.x, self.goal.y) not in came_from:
            return None

        path = []
        x, y = self.goal.x, self.goal.y
        while (x, y) != (self.start.x, self.start.y):
            path.append(PoseStamped(x=x, y=y))
            x, y = came_from[(x, y)]
        path.append(PoseStamped(x=self.start.x, y=self.start.y))
        path.reverse()
        return path

    def map_callback(self, msg):
        self.map = msg

    def start_callback(self, msg):
        self.start = msg.pose.position

    def goal_callback(self, msg):
        self.goal = msg.pose.position

if __name__ == '__main__':
    rospy.init_node('astar_planner')
    planner = AStarPlanner()
    rospy.Subscriber('/map', OccupancyGrid, planner.map_callback)
    rospy.Subscriber('/start', PoseStamped, planner.start_callback)
    rospy.Subscriber('/goal', PoseStamped, planner.goal_callback)
    path_pub = rospy.Publisher('/path', PoseStamped, queue_size=10)
    rate = rospy.Rate(10)
    while not rospy.is_shutdown():
        path = planner.plan()
        if path is not None:
            for pose in path:
                path_pub.publish(pose)
        rate.sleep()

在这个实现中，我们使用了一个优先队列来保存当前的搜索状态。我们还使用了一个字典来记录每个状态的父状态和当前的代价。在搜索过程中，我们逐步从优先队列中弹出状态，并计算它的所有邻居的代价。如果一个邻居是新的或者它的代价更小，我们就将它加入到优先队列中，并记录它的父状态和代价。当我们找到终点时，我们可以通过回溯父状态来重建路径。最后，我们将路径发布到ROS话题中。

这个实现是一个简单的A*算法，可能需要根据具体的使用情况进行优化。例如，可以使用更高效的数据结构来保存状态和代价，或者使用更复杂的启发式函数来加速搜索。