在一个100*100的区域内随机生成10个目标点，生成目标点之间的距离矩阵，利用麻雀搜索算法求出遍历每个目标点最短距离的目标函数 写出python代码

好的，以下是一个简单的 Python 实现：

import numpy as np

# 生成目标点
num_targets = 10
targets = np.random.rand(num_targets, 2) * 100

# 生成距离矩阵
distances = np.zeros((num_targets, num_targets))
for i in range(num_targets):
    for j in range(i+1, num_targets):
        distances[i, j] = np.linalg.norm(targets[i] - targets[j])
        distances[j, i] = distances[i, j]

# 定义目标函数
def fitness(solution):
    total_distance = 0
    for i in range(num_targets-1):
        total_distance += distances[solution[i], solution[i+1]]
    total_distance += distances[solution[-1], solution[0]]
    return total_distance

# 麻雀搜索算法
num_population = 20
max_generation = 100
mutation_rate = 0.1

population = np.zeros((num_population, num_targets), dtype=np.int)
for i in range(num_population):
    population[i] = np.random.permutation(num_targets)

for generation in range(max_generation):
    # 计算适应度
    fitness_values = np.array([fitness(solution) for solution in population])
    
    # 选择操作
    parents = np.zeros((num_population//2, num_targets), dtype=np.int)
    for i in range(num_population//2):
        idx = np.random.choice(num_population, size=2, replace=False)
        if fitness_values[idx[0]] < fitness_values[idx[1]]:
            parents[i] = population[idx[0]]
        else:
            parents[i] = population[idx[1]]
    
    # 交叉操作
    children = np.zeros((num_population//2, num_targets), dtype=np.int)
    for i in range(num_population//2):
        idx = np.random.choice(num_population//2, size=2, replace=False)
        cross_point = np.random.randint(num_targets-1)
        children[i, :cross_point+1] = parents[idx[0], :cross_point+1]
        for j in range(num_targets):
            if parents[idx[1], j] not in children[i]:
                children[i, cross_point+1] = parents[idx[1], j]
                cross_point += 1
                if cross_point == num_targets-1:
                    cross_point = 0
        children[i] = np.roll(children[i], np.random.randint(num_targets))
    
    # 变异操作
    for i in range(num_population//2):
        if np.random.rand() < mutation_rate:
            idx = np.random.choice(num_population//2)
            mutation_point = np.random.choice(num_targets, size=2, replace=False)
            children[i, mutation_point[0]], children[i, mutation_point[1]] = \
                children[i, mutation_point[1]], children[i, mutation_point[0]]
    
    # 更新种群
    population = np.concatenate((parents, children), axis=0)
    
    # 输出结果
    best_solution_idx = np.argmin(fitness_values)
    best_solution = population[best_solution_idx]
    print('Generation:', generation, 'Best Distance:', fitness_values[best_solution_idx], 'Best Solution:', best_solution)

注：该代码未进行参数调优，可能需要根据实际情况进行调整。