基于真实环境（经纬度和高度以及约束条件）下的无人机航路规划A星算法Python实现

以下是基于真实环境下的无人机航路规划A星算法的Python实现代码：

import math
import heapq

# 定义节点类
class Node:
    def __init__(self, x, y, z):
        self.x = x
        self.y = y
        self.z = z
        self.g = 0
        self.h = 0
        self.f = 0
        self.parent = None

    def __eq__(self, other):
        return self.x == other.x and self.y == other.y and self.z == other.z

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

    def __repr__(self):
        return f"({self.x}, {self.y}, {self.z})"


# 定义A星算法
def astar(start, goal, obstacles, resolution, max_altitude):
    # 计算两点之间距离
    def distance(n1, n2):
        dx = n1.x - n2.x
        dy = n1.y - n2.y
        dz = n1.z - n2.z
        return math.sqrt(dx ** 2 + dy ** 2 + dz ** 2)

    # 判断节点是否在障碍物内
    def in_obstacle(node, obstacles):
        for obstacle in obstacles:
            if obstacle[0] <= node.x <= obstacle[3] and \
                    obstacle[1] <= node.y <= obstacle[4] and \
                    obstacle[2] <= node.z <= obstacle[5]:
                return True
        return False

    # 判断节点是否越界
    def out_of_bounds(node, min_bound, max_bound):
        if node.x < min_bound[0] or node.y < min_bound[1] or node.z < min_bound[2] or \
                node.x > max_bound[0] or node.y > max_bound[1] or node.z > max_altitude:
            return True
        return False

    # 找到邻居节点
    def get_neighbors(node, obstacles, resolution):
        neighbors = []
        for dx in [-1, 0, 1]:
            for dy in [-1, 0, 1]:
                for dz in [-1, 0, 1]:
                    if dx == 0 and dy == 0 and dz == 0:
                        continue
                    x = node.x + dx * resolution
                    y = node.y + dy * resolution
                    z = node.z + dz * resolution
                    neighbor = Node(x, y, z)
                    if in_obstacle(neighbor, obstacles) or out_of_bounds(neighbor, min_bound, max_bound):
                        continue
                    neighbors.append(neighbor)
        return neighbors

    # 计算启发函数
    def heuristic(node, goal):
        dx = abs(node.x - goal.x)
        dy = abs(node.y - goal.y)
        dz = abs(node.z - goal.z)
        return math.sqrt(dx ** 2 + dy ** 2 + dz ** 2)

    # 初始化起点和终点
    start_node = Node(start[0], start[1], start[2])
    goal_node = Node(goal[0], goal[1], goal[2])

    # 设置搜索范围
    min_bound = (min(start[0], goal[0]), min(start[1], goal[1]), 0)
    max_bound = (max(start[0], goal[0]), max(start[1], goal[1]), max_altitude)

    # 初始化open列表和closed列表
    open_list = []
    closed_list = []

    # 将起点加入open列表
    heapq.heappush(open_list, start_node)

    # 开始搜索
    while len(open_list) > 0:
        # 取出open列表中f值最小的节点
        current_node = heapq.heappop(open_list)

        # 判断是否到达终点
        if current_node == goal_node:
            path = []
            while current_node is not None:
                path.append(current_node)
                current_node = current_node.parent
            return path[::-1]

        # 将当前节点加入closed列表
        closed_list.append(current_node)

        # 找到邻居节点并更新g、h、f值
        for neighbor in get_neighbors(current_node, obstacles, resolution):
            if neighbor in closed_list:
                continue
            tentative_g = current_node.g + distance(current_node, neighbor)
            if neighbor not in open_list:
                neighbor.h = heuristic(neighbor, goal_node)
                open_list.append(neighbor)
            elif tentative_g >= neighbor.g:
                continue
            neighbor.parent = current_node
            neighbor.g = tentative_g
            neighbor.f = neighbor.g + neighbor.h

    # 如果open列表为空，说明无法到达终点
    return None


# 测试代码
start = (0, 0, 0)
goal = (100, 100, 50)
obstacles = [(20, 30, 10, 40, 70, 30), (60, 50, 20, 80, 80, 40)]
resolution = 1
max_altitude = 50
path = astar(start, goal, obstacles, resolution, max_altitude)
if path:
    print("路径：")
    for node in path:
        print(node)
else:
    print("无法到达终点！")

注：以上代码仅作为参考，实际应用中需要根据具体情况进行修改和优化。