A*路径规划非常慢

使用优化算法，如JPS（Jump Point Search）或Theta *算法，来加速A *的执行速度。

示例代码：

JPS实现：

def jump(point, direction, end_point)
    # 跳跃到下一个相邻节点
    next_point = point + direction

    if not end_point.encompasses(next_point):
        # 当前位置不在终点范围内，终止跳跃
        return None

    if next_point == end_point:
        # 找到了终点
        return next_point

    if self.grid.is_obstacle(next_point):
        # 下一点是障碍，终止跳跃
        return None

    # 当前节点沿目标线方向跳跃
    if direction.is_diagonal():
        if any((
            self.jump(next_point, Direction(direction.x, 0), end_point),
            self.jump(next_point, Direction(0, direction.y), end_point)
        )):
            # 投影到x, y轴，是否有可达节点
            return next_point
        return None
    else:
        return self.jump(next_point, direction, end_point)

def get_successors(current_node, end_node):
    successors = []
    # 寻找可到达的节点，如果找到加进successors
    for direction in Direction.diagonal_directions() + Direction.cardinal_directions():
        neighbor = jump(current_node.position, direction, end_node.position)
        if neighbor is None:
            continue
        cost = current_node.g + current_node.position.distance_to(neighbor) # 计算代价g
        successor = Node(neighbor, current_node, cost, neighbor.distance_to(end_node.position)) # 计算代价f
        successors.append(successor)
    return successors

Theta *实现：

def walk(current_node, end_node):
    direction = current_node.position.direction_to(end_node.position)
    if direction is None or not self.line_of_sight(current_node.position, end_node.position):
        return None
    if current_node.position == end_node.position:
        return end_node
    neighbor_pos = current_node.position + direction
    neighbor = self.grid.node_at(neighbor_pos)
    # 计算代价
    cost = current_node.g + current_node.position.distance_to(neighbor_pos)
    successor = Node(neighbor_pos, current_node, cost, neighbor_pos.distance_to(end_node.position))
    return successor

def get_successors(current_node, end_node):
    successors = []
    # 寻找可到达的节点，如果找到加进successors