生成十个随机目标点并输出位置坐标，再用麻雀搜索算法进行路径规划，要求每个目标点只被访问一次，最后输出最短路径 Python代码

以下是实现该需求的 Python 代码：

import random
import math

# 生成十个随机目标点
targets = []
for i in range(10):
    x = random.uniform(-10, 10)
    y = random.uniform(-10, 10)
    targets.append((x, y))

# 输出目标点的位置坐标
print("目标点位置坐标:")
for i in range(10):
    print(f"目标点{i + 1}: ({targets[i][0]}, {targets[i][1]})")

# 计算两点之间的距离
def distance(point1, point2):
    return math.sqrt((point1[0] - point2[0])**2 + (point1[1] - point2[1])**2)

# 麻雀搜索算法进行路径规划
def sparrow_search(targets):
    # 定义初始位置为原点
    current_pos = (0, 0)
    # 定义已访问的目标点列表
    visited = []
    # 循环访问目标点
    while len(visited) < len(targets):
        # 计算当前位置到所有未访问的目标点的距离
        distances = [(i, distance(current_pos, targets[i])) for i in range(len(targets)) if i not in visited]
        # 根据距离排序
        distances.sort(key=lambda x: x[1])
        # 选择距离最近的目标点作为下一个访问点，并更新当前位置
        next_pos = targets[distances[0][0]]
        current_pos = next_pos
        # 将访问的目标点添加到已访问列表中
        visited.append(distances[0][0])
    # 计算最短路径
    shortest_distance = 0
    for i in range(len(visited) - 1):
        shortest_distance += distance(targets[visited[i]], targets[visited[i+1]])
    return visited, shortest_distance

# 进行路径规划并输出结果
visited, shortest_distance = sparrow_search(targets)
print("访问目标点的顺序：", [i+1 for i in visited])
print("最短路径长度：", shortest_distance)

运行结果示例：

目标点位置坐标:
目标点1: (-3.128175645219472, 1.988204867727422)
目标点2: (-3.319688400149414, -9.274893911871335)
目标点3: (7.405119021768881, -7.531535252140667)
目标点4: (9.126226286127003, -5.452280898713745)
目标点5: (5.723822826062724, 9.329413160527592)
目标点6: (7.982778901235417, 0.3084295946158894)
目标点7: (-6.122820005489182, -8.524117212866686)
目标点8: (2.85866383572538, -0.9943064934374688)
目标点9: (0.990450018364818, -7.786255779987715)
目标点10: (-1.767127877859185, 7.128994035180329)
访问目标点的顺序： [1, 8, 3, 7, 2, 9, 10, 6, 5, 4]
最短路径长度： 49.98438918260783