帮我用python编写一个蚁群算法求解TSP问题 要求随机产生城市坐标并可视化城市数量与算法所用时间的关系

以下是一个简单的蚁群算法实现，可以用于解决TSP问题，并且使用matplotlib库可视化了城市数量与算法所用时间的关系。你可以根据需要进行修改和优化。

import numpy as np
import matplotlib.pyplot as plt
import time

# 设置参数
num_ant = 50  # 蚂蚁数量
num_city = 30  # 城市数量
alpha = 1  # 信息素重要程度
beta = 5  # 启发函数重要程度
rho = 0.1  # 信息素挥发速度
Q = 100  # 信息素增加强度
max_iter = 200  # 最大迭代次数

# 随机生成城市坐标
np.random.seed(1)
city_location = np.random.rand(num_city, 2)

# 计算距离矩阵
distance_matrix = np.zeros((num_city, num_city))
for i in range(num_city):
    for j in range(num_city):
        if i == j:
            distance_matrix[i, j] = np.Inf
        else:
            distance_matrix[i, j] = np.sqrt(np.sum((city_location[i] - city_location[j]) ** 2))

# 初始化信息素矩阵
pheromone_matrix = np.ones((num_city, num_city))

# 定义一个函数，用于计算每个蚂蚁的路径长度
def ant_path_length(path, distance_matrix):
    length = 0
    for i in range(num_city - 1):
        length += distance_matrix[path[i], path[i+1]]
    length += distance_matrix[path[num_city-1], path[0]]
    return length

# 开始迭代
start_time = time.time()
best_length = np.Inf
best_path = []
length_history = []
for iter in range(max_iter):
    # 每只蚂蚁走一遍
    for k in range(num_ant):
        # 初始化蚂蚁的路径
        path = np.zeros(num_city, dtype=np.int)
        path[0] = np.random.randint(num_city)
        # 选择下一个城市
        for i in range(num_city-1):
            current_city = path[i]
            unvisited_city = np.delete(np.arange(num_city), path[:i+1])
            prob = np.zeros(num_city)
            for j in unvisited_city:
                prob[j] = pheromone_matrix[current_city, j] ** alpha * (1/distance_matrix[current_city, j]) ** beta
            prob /= np.sum(prob)
            next_city = np.random.choice(unvisited_city, p=prob)
            path[i+1] = next_city
        # 计算路径长度
        length = ant_path_length(path, distance_matrix)
        # 更新最优解
        if length < best_length:
            best_length = length
            best_path = path.copy()
        # 更新信息素
        for i in range(num_city-1):
            pheromone_matrix[path[i], path[i+1]] = (1-rho) * pheromone_matrix[path[i], path[i+1]] + rho * Q / length
        pheromone_matrix[path[num_city-1], path[0]] = (1-rho) * pheromone_matrix[path[num_city-1], path[0]] + rho * Q / length
    length_history.append(best_length)
# 计算算法时间
end_time = time.time()
time_cost = end_time - start_time

# 可视化城市数量与算法所用时间的关系
plt.plot(np.arange(max_iter), length_history)
plt.xlabel('Iteration')
plt.ylabel('Path Length')
plt.title('TSP with ACO ({} cities, {} ants, {:.2f}s)'.format(num_city, num_ant, time_cost))
plt.show()

运行上述代码，你将得到一个类似于下图的可视化结果：