用matlab程序写出使用人工蜂群算法实现的0-1背包问题的代码

以下是使用人工蜂群算法实现的0-1背包问题的Matlab代码：

% 0-1背包问题
% 使用人工蜂群算法进行求解
clc;
clear;
close all;

% 参数设置
n = 20; % 物品数量
c = 50; % 背包容量
w = randi([1, 10], [1, n]); % 物品重量
v = randi([10, 100], [1, n]); % 物品价值
max_iter = 100; % 最大迭代次数
n_bees = 50; % 蜜蜂数量
limit = 20; % 限制值

% 初始化
x = zeros(n, 1); % 物品选择向量
L = zeros(n_bees, 1); % 蜜蜂距离
P = zeros(n_bees, n); % 蜜蜂位置
F = zeros(n_bees, 1); % 蜜蜂适应度
best_solution = []; % 最佳解
best_fitness = 0; % 最佳适应度

% 计算所有蜜蜂的适应度
for i = 1 : n_bees
    x = randi([0, 1], [n, 1]);
    W = w * x;
    if W > c % 如果超过了容量，则强制将一些物品移除
        for j = 1 : limit
            idx = find(x == 1, 1);
            x(idx) = 0;
            W = w * x;
            if W <= c
                break;
            end
        end
    end
    V = v * x;
    F(i) = V;
    P(i, :) = x';
    % 更新最佳解
    if V > best_fitness
        best_fitness = V;
        best_solution = x;
    end
end

% 迭代优化
for t = 1 : max_iter
    % 采蜜蜂
    for i = 1 : n_bees
        x = P(i, :);
        j = randi([1, n]);
        k = randi([1, n]);
        while k == j
            k = randi([1, n]);
        end
        x(j) = 1 - x(j);
        W = w * x';
        if W > c % 如果超过了容量，则强制将一些物品移除
            for j = 1 : limit
                idx = find(x == 1, 1);
                x(idx) = 0;
                W = w * x';
                if W <= c
                    break;
                end
            end
        end
        V = v * x';
        L(i) = abs(F(i) - V);
        % 如果新的位置更好，则更新
        if V > F(i)
            F(i) = V;
            P(i, :) = x;
        end
    end
    
    % 侦查蜂
    for i = 1 : n_bees
        if rand() < 0.1 % 10%的概率进行随机搜索
            x = randi([0, 1], [n, 1]);
            W = w * x;
            if W > c % 如果超过了容量，则强制将一些物品移除
                for j = 1 : limit
                    idx = find(x == 1, 1);
                    x(idx) = 0;
                    W = w * x;
                    if W <= c
                        break;
                    end
                end
            end
            V = v * x;
            if V > F(i)
                F(i) = V;
                P(i, :) = x';
            end
        end
    end
    
    % 工蜂
    for i = 1 : n_bees
        x = P(i, :);
        j = randi([1, n]);
        k = randi([1, n]);
        while k == j
            k = randi([1, n]);
        end
        x(j) = 1 - x(j);
        W = w * x';
        if W > c % 如果超过了容量，则强制将一些物品移除
            for j = 1 : limit
                idx = find(x == 1, 1);
                x(idx) = 0;
                W = w * x';
                if W <= c
                    break;
                end
            end
        end
        V = v * x';
        L(i) = abs(F(i) - V);
        % 如果新的位置更好，则更新
        if V > F(i)
            F(i) = V;
            P(i, :) = x;
        end
    end
    
    % 计算所有蜜蜂的适应度
    for i = 1 : n_bees
        x = P(i, :);
        W = w * x';
        if W > c % 如果超过了容量，则强制将一些物品移除
            for j = 1 : limit
                idx = find(x == 1, 1);
                x(idx) = 0;
                W = w * x';
                if W <= c
                    break;
                end
            end
        end
        V = v * x';
        F(i) = V;
        P(i, :) = x';
        % 更新最佳解
        if V > best_fitness
            best_fitness = V;
            best_solution = x;
        end
    end
    
    % 显示当前最优解
    fprintf('Iteration: %d  Best Solution: %d\n', t, best_fitness);
end

fprintf('\nBest Solution:\n');
disp(best_solution');
fprintf('Best Fitness: %d\n', best_fitness);