多旅行商问题matlab代码

以下是一个简单的多旅行商问题的 MATLAB 代码，使用遗传算法进行解决：

%% 多旅行商问题 - 遗传算法
clc;
clear;
close all;

% 问题参数
numCities = 50;         % 城市数目
numSalesmen = 5;        % 旅行商数目
mapSize = 100;          % 地图大小
mutationRate = 0.05;    % 变异率
numGenerations = 200;   % 迭代次数

% 随机生成城市坐标
cities = rand(numCities, 2) * mapSize;

% 初始化种群
population = zeros(numCities, numSalesmen, numGenerations);
for i = 1:numGenerations
    % 每个旅行商随机分配一个城市
    population(:, :, i) = randperm(numCities);
end

% 开始迭代
for i = 1:numGenerations
    % 计算每个个体的适应度，即总旅行距离
    fitness = zeros(1, size(population, 3));
    for j = 1:size(population, 3)
        dist = 0;
        for k = 1:numSalesmen
            tour = population(:, k, j);
            dist = dist + tourDistance(cities(tour, :));
        end
        fitness(j) = dist;
    end
    
    % 选择操作
    [~, idx] = sort(fitness);
    elite = population(:, :, idx(1:2));
    selectionProb = fitness / sum(fitness);
    parents = zeros(size(population));
    for j = 1:size(population, 3)
        parents(:, :, j) = rouletteWheelSelection(selectionProb);
    end
    
    % 交叉操作
    children = crossover(parents);
    
    % 变异操作
    children = mutation(children, mutationRate);
    
    % 更新种群
    population(:, :, 1:i) = elite;
    population(:, :, i+1:end) = children;
    
    % 绘制当前最优解
    [~, idx] = min(fitness);
    bestTour = population(:, :, idx);
    bestDist = fitness(idx);
    clf;
    plotTsp(cities, bestTour, bestDist);
    drawnow;
end

% 绘制旅行路线
plotTsp(cities, bestTour, bestDist);

% 计算旅行距离
function dist = tourDistance(tour)
    n = size(tour, 1);
    dist = sum(sqrt(sum((tour - circshift(tour, 1)).^2, 2)));
end

% 轮盘赌选择
function parents = rouletteWheelSelection(selectionProb)
    n = size(selectionProb, 2);
    parents = zeros(size(selectionProb));
    for i = 1:size(parents, 2)
        idx = find(rand <= cumsum(selectionProb), 1);
        parents(:, i) = population(:, :, idx);
    end
end

% 交叉操作
function children = crossover(parents)
    numChildren = size(parents, 3);
    children = zeros(size(parents));
    for i = 1:numChildren
        % 选择两个父代
        p1 = parents(:, :, randi(size(parents, 3)));
        p2 = parents(:, :, randi(size(parents, 3)));
        
        % 选择一个交叉点
        crossPoint = randi(size(parents, 1));
        
        % 交叉操作
        child1 = [p1(1:crossPoint, :); p2(crossPoint+1:end, :)];
        child2 = [p2(1:crossPoint, :); p1(crossPoint+1:end, :)];
        
        % 随机选择一个子代
        if rand < 0.5
            children(:, :, i) = child1;
        else
            children(:, :, i) = child2;
        end
    end
end

% 变异操作
function children = mutation(parents, mutationRate)
    numMutations = floor(size(parents, 1) * mutationRate);
    children = parents;
    for i = 1:size(parents, 3)
        % 随机选择需要变异的子代
        idx = randperm(size(parents, 1), numMutations);
        for j = 1:numMutations
            % 选择两个需要交换的城市
            swapIdx = randperm(size(parents, 2), 2);
            children(idx(j), swapIdx(1), i) = parents(idx(j), swapIdx(2), i);
            children(idx(j), swapIdx(2), i) = parents(idx(j), swapIdx(1), i);
        end
    end
end

% 绘制旅行路线图
function plotTsp(cities, tour, dist)
    hold on;
    set(gca, 'xtick', [], 'ytick', []);
    plot(cities(:, 1), cities(:, 2), 'ko');
    for i = 1:size(tour, 2)
        plot(cities(tour(:, i), 1), cities(tour(:, i), 2), 'color', rand(1, 3));
    end
    title(sprintf('旅行距离: %0.2f', dist));
end

这段代码使用遗传算法进行解决多旅行商问题。首先，随机生成指定数目的城市坐标。然后，使用遗传算法进行迭代求解，每次迭代包括以下步骤：

- 计算每个个体的适应度，即总旅行距离；
- 根据适应度计算选择概率，使用轮盘赌选择算法选择父代；
- 使用交叉操作生成子代；
- 使用变异操作对子代进行变异；
- 更新种群，包括保留精英个体和添加新的子代；
- 绘制当前最优解。

最终，程序输出旅行路线图和旅行距离。注意，这个程序只是一个简单的多旅行商问题的求解方法，仅供参考。如果需要更高效的算法或更复杂的问题求解，请参考相关文献或其他资料。