有十个点横坐标分别为7 10 14 17 20 20 23 23 27 30纵坐标为25 73 54 2 51 31 31 79 79 28货物需求量分别为2 2 3 5 4 9 6 10 6 3使用禁忌搜索算法使用MATLAB实现总里程数最短的目标为了简化问题限制每个顾客只能被分配到一条路径一共有十辆车每辆车的容量是10

以下是MATLAB代码实现：

clear; clc;

%初始化数据 x = [7 10 14 17 20 20 23 23 27 30]; y = [25 73 54 2 51 31 31 79 79 28]; demand = [2 2 3 5 4 9 6 10 6 3]; num_customers = length(x); num_vehicles = 10; vehicle_capacity = 10; max_iterations = 1000;

%计算距离矩阵 dist_matrix = zeros(num_customers,num_customers); for i = 1:num_customers for j = 1:num_customers dist_matrix(i,j) = sqrt((x(i)-x(j))^2 + (y(i)-y(j))^2); end end

%初始化禁忌表 tabu_list = zeros(num_customers,num_customers);

%初始化解 solution = zeros(num_vehicles,num_customers); for i = 1:num_vehicles solution(i,1) = 1; solution(i,end) = 1; end

%计算初始解的总里程数 total_distance = calculate_distance(solution,dist_matrix);

%记录最优解 best_solution = solution; best_distance = total_distance;

%禁忌搜索算法 iteration = 1; while iteration <= max_iterations %找到当前解中需要改变的边 [i,j] = find_edge_to_swap(solution,vehicle_capacity,demand,tabu_list);

%交换边
new_solution = swap_edge(solution,i,j);

%计算新解的总里程数
new_distance = calculate_distance(new_solution,dist_matrix);

%更新禁忌表
tabu_list(i,j) = iteration + 10;

%更新当前解
solution = new_solution;
total_distance = new_distance;

%如果新解更优，则更新最优解
if total_distance < best_distance
    best_solution = new_solution;
    best_distance = total_distance;
end

%迭代次数加一
iteration = iteration + 1;

end

%输出结果 disp("最短总里程数为：" + best_distance); for i = 1:num_vehicles route = find(best_solution(i,:)==1); disp("第" + i + "辆车的路径为：" + num2str(route)); end

%计算解的总里程数的函数 function total_distance = calculate_distance(solution,dist_matrix) total_distance = 0; for i = 1:size(solution,1) for j = 1:size(solution,2)-1 if solution(i,j) == 1 total_distance = total_distance + dist_matrix(j,solution(i,j+1)); end end end end

%找到需要交换的边的函数 function [i,j] = find_edge_to_swap(solution,vehicle_capacity,demand,tabu_list) num_vehicles = size(solution,1); num_customers = size(solution,2); best_i = 0; best_j = 0; best_delta_distance = Inf; for k = 1:num_vehicles for i = 2:num_customers-1 if solution(k,i) == 1 for j = i+1:num_customers-1 if solution(k,j) == 0 && demand(j-1) <= vehicle_capacity delta_distance = calculate_delta_distance(solution,k,i,j); if delta_distance < best_delta_distance && tabu_list(i,j) == 0 best_i = i; best_j = j; best_delta_distance = delta_distance; end end end end end end i = best_i; j = best_j; end

%计算交换边后的总里程数变化的函数 function delta_distance = calculate_delta_distance(solution,k,i,j) num_customers = size(solution,2); route = find(solution(k,:)==1); distance1 = 0; for l = 1:length(route)-1 distance1 = distance1 + dist_matrix(route(l),route(l+1)); end distance1 = distance1 - dist_matrix(route(i-1),route(i)) - dist_matrix(route(j),route(j+1)) + dist_matrix(route(i-1),route(j)) + dist_matrix(route(i),route(j+1)); distance2 = 0; new_route = [route(1:i-1) route(j) route(i+1:j-1) route(i) route(j+1:end)]; for l = 1:length(new_route)-1 distance2 = distance2 + dist_matrix(new_route(l),new_route(l+1)); end delta_distance = distance2 - distance1; end

%交换边的函数 function new_solution = swap_edge(solution,i,j) new_solution = solution; for k = 1:size(solution,1) if solution(k,i) == 1 new_solution(k,i) = 0; new_solution(k,j) = 1; end end end