3维多旅行商matlab实现
时间: 2023-08-15 17:18:54 浏览: 42
三维多旅行商问题是一个NP难问题,没有确定的多项式时间算法。因此,使用穷举法或者遗传算法等启发式算法进行求解。以下是使用遗传算法求解三维多旅行商问题的 MATLAB 实现示例:
```matlab
% 定义旅行商数量 n 和城市数量 m
n = 3;
m = 5;
% 随机生成城市坐标
cities = rand(m, 3);
% 计算城市之间的距离矩阵
dist_mat = zeros(m, m);
for i = 1:m
for j = 1:m
dist_mat(i,j) = norm(cities(i,:) - cities(j,:));
end
end
% 遗传算法参数设置
pop_size = 100; % 种群大小
num_iter = 1000; % 迭代次数
mutation_rate = 0.02; % 变异率
% 初始化种群
pop = zeros(pop_size, m * n);
for i = 1:pop_size
pop(i,:) = randperm(m * n);
end
% 开始迭代
for iter = 1:num_iter
% 计算适应度函数
fitness = zeros(pop_size, 1);
for i = 1:pop_size
tour = reshape(pop(i,:), n, m);
tour_length = 0;
for j = 1:n
tour_length = tour_length + sum(dist_mat(tour(j,:), circshift(tour(j,:), -1)));
end
fitness(i) = 1 / tour_length;
end
% 选择操作
idx = randsample(pop_size, pop_size, true, fitness);
pop = pop(idx,:);
% 交叉操作
for i = 1:2:pop_size
parent1 = pop(i,:);
parent2 = pop(i+1,:);
crossover_point = randi([1, m*n-1]);
child1 = [parent1(1:crossover_point) parent2(crossover_point+1:end)];
child2 = [parent2(1:crossover_point) parent1(crossover_point+1:end)];
pop(i,:) = child1;
pop(i+1,:) = child2;
end
% 变异操作
for i = 1:pop_size
if rand < mutation_rate
idx1 = randi([1, m*n]);
idx2 = randi([1, m*n]);
pop(i,[idx1 idx2]) = pop(i,[idx2 idx1]);
end
end
% 输出当前迭代结果
best_tours = zeros(n, m);
best_fitness = 0;
for i = 1:n
[val,idx] = max(fitness);
best_tours(i,:) = reshape(pop(idx,:), m, n)';
best_fitness = best_fitness + val;
fitness(idx) = -Inf;
end
disp(['Iteration ', num2str(iter), ': best fitness = ', num2str(best_fitness/n)]);
end
% 输出最终结果
disp('Best tours:');
disp(best_tours);
disp(['Average tour length = ', num2str(1/(best_fitness/n))]);
```
该代码使用遗传算法求解三维多旅行商问题,具体实现细节可以参考代码注释。注意,由于遗传算法是随机化算法,因此不同运行结果可能会有所不同。