写一个Matlab从配送中心出发多条线路连接其它点，每条线路不可连接超过6个节点并使路线长度之和最短

您的问题是一个典型的带约束条件的多旅行商问题（Constrained Multiple Traveling Salesman Problem，CMTSP）。该问题也是NP难问题，通常使用启发式算法求解。以下是一个使用遗传算法（Genetic Algorithm，GA）求解CMTSP的Matlab代码示例：

% 配置参数
popSize = 50; % 种群大小
numCities = 30; % 城市数
maxGen = 500; % 进化代数
pc = 0.8; % 交叉概率
pm = 0.01; % 变异概率
maxNodes = 6; % 每条线路最大节点数

% 生成随机城市坐标
cities = rand(numCities, 2) * 100;

% 初始化种群
numRoutes = ceil(numCities/maxNodes);
pop = zeros(numCities+numRoutes-1, popSize);
for i = 1:popSize
    idx = randperm(numCities);
    for j = 1:numRoutes
        startIdx = (j-1)*maxNodes + 1;
        endIdx = min(j*maxNodes, numCities);
        pop(startIdx:endIdx+(j<numRoutes), i) = idx(startIdx:endIdx);
    end
end

% 进化过程
for gen = 1:maxGen
    % 计算适应度值
    dist = zeros(popSize, 1);
    for i = 1:popSize
        routes = cell(numRoutes, 1);
        for j = 1:numRoutes
            startIdx = (j-1)*maxNodes + 1;
            endIdx = min(j*maxNodes, numCities);
            route = pop(startIdx:endIdx+(j<numRoutes), i);
            route(route==0) = [];
            routes{j} = route;
        end
        
        % 计算路线长度
        d = 0;
        for j = 1:numRoutes
            route = routes{j};
            d = d + sum(sqrt(sum((cities(route, :) - circshift(cities(route, :), [1, 0])).^2, 2)));
        end
        dist(i) = d;
    end
    fitness = 1./dist;
    fitness = fitness./sum(fitness);

    % 种群选择
    [val, idx] = sort(fitness, 'descend');
    elite = pop(:, idx(1));
    pop = pop(:, idx(1:popSize-1));
    prob = cumsum(val);
    prob = prob/prob(end);
    for i = 1:popSize-1
        a = find(rand() <= prob, 1);
        b = find(rand() <= prob, 1);
        if rand() <= pc
            % 交叉
            c = randi(numRoutes-1);
            routeA = cell(numRoutes, 1);
            routeB = cell(numRoutes, 1);
            for j = 1:numRoutes
                if j <= c
                    routeA{j} = routes{a}{j};
                    routeB{j} = routes{b}{j};
                else
                    routeA{j} = routes{b}{j};
                    routeB{j} = routes{a}{j};
                end
            end
            childA = [];
            childB = [];
            for j = 1:numRoutes
                route = routeA{j};
                for k = 1:length(route)
                    if sum(ismember(routeB{j}, route(k))) == 0
                        childA(end+1) = route(k);
                    end
                end
                missing = setdiff(routeB{j}, childA);
                if length(missing) > 0
                    childA(end+1:end+length(missing)) = missing;
                end
                
                route = routeB{j};
                for k = 1:length(route)
                    if sum(ismember(routeA{j}, route(k))) == 0
                        childB(end+1) = route(k);
                    end
                end
                missing = setdiff(routeA{j}, childB);
                if length(missing) > 0
                    childB(end+1:end+length(missing)) = missing;
                end
                
                % 控制每条线路的节点数
                if length(childA) > maxNodes
                    childA = childA(1:maxNodes);
                end
                if length(childB) > maxNodes
                    childB = childB(1:maxNodes);
                end
            end
            childA = [childA, zeros(1, numCities-length(childA))];
            childB = [childB, zeros(1, numCities-length(childB))];
            pop(:, i) = childA;
            pop(:, i+1) = childB;
        else
            % 复制
            pop(:, i) = pop(:, a);
        end
    end

    % 变异
    for i = 2:popSize-1
        if rand() <= pm
            a = randi(numRoutes);
            b = randi(numRoutes);
            route = routes{a};
            pos = randperm(length(route), 2);
            route([pos(1), pos(2)]) = route([pos(2), pos(1)]);
            if length(route) > maxNodes
                route = route(1:maxNodes);
            end
            routes{a} = route;
            pop(:, i) = zeros(numCities+numRoutes-1, 1);
            for j = 1:numRoutes
                startIdx = (j-1)*maxNodes + 1;
                endIdx = min(j*maxNodes, numCities);
                if j == a
                    pop(startIdx:endIdx+(j<numRoutes), i) = route;
                else
                    pop(startIdx:endIdx+(j<numRoutes), i) = routes{j};
                end
            end
        end
    end
    
    % 加入精英
    pop(:, end) = elite;
end

% 输出最优解
routes = cell(numRoutes, 1);
for j = 1:numRoutes
    startIdx = (j-1)*maxNodes + 1;
    endIdx = min(j*maxNodes, numCities);
    route = pop(startIdx:endIdx+(j<numRoutes), idx(1));
    route(route==0) = [];
    routes{j} = route;
end
d = 0;
for j = 1:numRoutes
    route = routes{j};
    d = d + sum(sqrt(sum((cities(route, :) - circshift(cities(route, :), [1, 0])).^2, 2)));
end
fprintf('最短路程长度：%f\n', d);
fprintf('最短路程路径：\n');
for j = 1:numRoutes
    route = routes{j};
    fprintf('[%d, %d, %d, %d, %d, %d]\n', route);
end

该代码使用遗传算法求解CMTSP问题，将城市坐标随机生成，通过进化过程逐步优化种群中的路线，并控制每条线路的节点数不超过6个。最终输出最短路程长度和路径。您可以根据实际情况进行参数调整和优化。