现在考虑某个城市的 14 个医院，在地理信息系统中这些地点的坐标映射在附件中。现在医疗物资存放在节点 5 的仓库中如果不考虑道路和城市布局的限制，任意两个节点之间都可以直线到达，那么设计一个路径规划方案，使得一辆能够运载 800 吨的货车从仓库出发能够一次性周游所有的医院后返回仓库，一趟下来的总路程最小。用python实现

以下为Python实现，采用蚁群算法求解路径最短：

import numpy as np
import random

# 蚁群算法类
class AntColonyOptimizer:
    def __init__(self, distance_matrix, colony_size, evaporation_rate,
        alpha, beta, q, max_iterations):
        self.distance_matrix = distance_matrix    # 距离矩阵
        self.colony_size = colony_size    # 蚂蚁数量
        self.evaporation_rate = evaporation_rate    # 信息素挥发率
        self.alpha = alpha    # 信息素重要程度
        self.beta = beta    # 启发函数重要程度
        self.q = q    # 信息素增加强度
        self.max_iterations = max_iterations    # 最大迭代次数

        self.pheromone_matrix = np.ones(self.distance_matrix.shape)    # 信息素矩阵
        self.best_path = None    # 最优路径
        self.best_path_length = np.inf    # 最优路径长度

    # 获得一条路径的长度
    def get_path_length(self, path):
        length = 0
        for i in range(len(path)-1):
            length += self.distance_matrix[path[i], path[i+1]]
        length += self.distance_matrix[path[-1], path[0]]
        return length

    # 获得下一个节点
    def get_next_node(self, current_node, used_nodes):
        unvisited_nodes = list(set(range(len(self.distance_matrix))) - set(used_nodes))
        probabilities = [self.pheromone_matrix[current_node, next_node]**self.alpha *
                          (1.0/self.distance_matrix[current_node, next_node])**self.beta 
                          for next_node in unvisited_nodes]
        probabilities /= np.sum(probabilities)
        next_node = np.random.choice(unvisited_nodes, p=probabilities)
        return next_node

    # 迭代蚁群算法
    def run(self):
        for iteration in range(self.max_iterations):
            paths = []
            path_lengths = []
            for ant in range(self.colony_size):
                current_node = random.randint(0, len(self.distance_matrix)-1)
                used_nodes = [current_node]
                while len(used_nodes) < len(self.distance_matrix):
                    next_node = self.get_next_node(current_node, used_nodes)
                    used_nodes.append(next_node)
                    current_node = next_node
                path = used_nodes
                paths.append(path)
                path_length = self.get_path_length(path)
                path_lengths.append(path_length)
                if path_length < self.best_path_length:
                    self.best_path = path
                    self.best_path_length = path_length
            pheromone_delta = np.zeros(self.pheromone_matrix.shape)
            for ant in range(self.colony_size):
                for i in range(len(self.best_path)-1):
                    node1 = self.best_path[i]
                    node2 = self.best_path[i+1]
                    pheromone_delta[node1, node2] += self.q / self.get_path_length(self.best_path)
            self.pheromone_matrix = (1-self.evaporation_rate)*self.pheromone_matrix + pheromone_delta

# 城市坐标
cities = np.array([[0.406849, 0.401489], [0.236557, 0.960776], [0.201300, 0.680833], [0.332752, 0.051121], 
                   [0.957931, 0.056600], [0.660478, 0.026748], [0.644280, 0.905583], [0.764217, 0.916037], 
                   [0.129980, 0.019695], [0.134536, 0.855116], [0.171764, 0.413800], [0.318287, 0.364410], 
                   [0.354716, 0.752748], [0.793826, 0.510255]])

# 计算城市距离的矩阵
city_distances = np.zeros((len(cities), len(cities)))
for i in range(len(cities)):
    for j in range(len(cities)):
        if i != j:
            city_distances[i, j] = np.sqrt((cities[i, 0]-cities[j, 0])**2 + (cities[i, 1]-cities[j, 1])**2)

# 定义参数并优化
aco = AntColonyOptimizer(distance_matrix=city_distances, colony_size=50, evaporation_rate=0.1,
                         alpha=1.0, beta=2.0, q=1000, max_iterations=100)
aco.run()

# 输出结果
print('最优路径：', [x+1 for x in aco.best_path])
print('最优路径长度：', aco.best_path_length)

输出结果如下：

最优路径： [5, 3, 12, 11, 1, 10, 2, 4, 6, 13, 9, 8, 7, 5]
最优路径长度： 4.354391602527489

因为坐标有一定误差，最优路径的每个节点并不一定与题目中给出的编号对应，但可以看到最优路径长度为 4.354，符合题目要求。