给出背包问题分支定界算法的MATLAB代码

function [maxValue, optimalItems] = branchAndBound(capacity, weights, values) % 计算分支定界算法的背包问题解决方案 % 输入： % capacity - 背包容量 % weights - 物品的重量向量 % values - 物品的价值向量 % 输出： % maxValue - 最大价值 % optimalItems - 最优解的物品向量

n = length(weights); % 物品数量 maxValue = -1; % 最大价值初值 optimalItems = zeros(1, n); % 最优解初值

% 初始化根节点 node.level = 0; % 节点层数 node.bound = bound(node, capacity, weights, values, n); % 上界 node.weight = 0; % 节点的重量 node.value = 0; % 节点的价值 node.taken = zeros(1, n); % 节点的物品向量 queue = PriorityQueue(); % 创建优先队列 queue.insert(node, -node.bound); % 将根节点加入优先队列

% 进入主循环 while ~queue.isEmpty() % 取出优先队列中优先级最高的节点 node = queue.pop(); % 如果该节点的上界小于当前最大价值，剪枝 if node.bound <= maxValue continue; end % 如果该节点是叶节点，更新最大价值和最优解 if node.level == n if node.value > maxValue maxValue = node.value; optimalItems = node.taken; end else % 展开非叶节点 % 情况1：不选该节点的下一个物品 child = node; child.level = node.level + 1; child.bound = bound(child, capacity, weights, values, n); queue.insert(child, -child.bound); % 情况2：选择该节点的下一个物品 if node.weight + weights(child.level) <= capacity child = node; child.level = node.level + 1; child.weight = node.weight + weights(child.level); child.value = node.value + values(child.level); child.taken(child.level) = 1; child.bound = bound(child, capacity, weights, values, n); queue.insert(child, -child.bound); end end end

end

function b = bound(node, capacity, weights, values, n) % 计算节点的上界 b = node.value; w = node.weight; j = node.level + 1; while j <= n && w + weights(j) <= capacity b = b + values(j); w = w + weights(j); j = j + 1; end if j <= n b = b + (capacity - w) * values(j) / weights(j); end end

% 优先队列类 classdef PriorityQueue < handle properties (Access = private) heap % 堆 priorities % 堆中元素的优先级 count % 堆中元素数量 end

methods
    function obj = PriorityQueue()
        % 构造函数
        obj.heap = {};
        obj.priorities = [];
        obj.count = 0;
    end
    
    function insert(obj, item, priority)
        % 插入元素到优先队列中
        obj.count = obj.count + 1;
        obj.heap{obj.count} = item;
        obj.priorities(obj.count) = priority;
        obj.fixUp(obj.count);
    end
    
    function item = pop(obj)
        % 从优先队列中取出优先级最高的元素，并删除该元素
        item = obj.heap{1};
        obj.heap{1} = obj.heap{obj.count};
        obj.priorities(1) = obj.priorities(obj.count);
        obj.count = obj.count - 1;
        obj.fixDown(1);
    end
    
    function f = isEmpty(obj)
        % 判断优先队列是否为空
        f = obj.count == 0;
    end
end

methods (Access = private)
    function fixUp(obj, k)
        % 将元素上移，使得堆保持最大堆性质
        while k > 1 && obj.priorities(k) > obj.priorities(obj.parent(k))
            obj.swap(k, obj.parent(k));
            k = obj.parent(k);
        end
    end
    
    function fixDown(obj, k)
        % 将元素下移，使得堆保持最大堆性质
        while obj.leftChild(k) <= obj.count
            child = obj.leftChild(k);
            if child < obj.count && obj.priorities(child) < obj.priorities(child + 1)
                child = child + 1;
            end
            if obj.priorities(k) >= obj.priorities(child)
                break;
            end
            obj.swap(k, child);
            k = child;
        end
    end
    
    function swap(obj, i, j)
        % 交换堆中两个元素
        temp = obj.heap{i};
        obj.heap{i} = obj.heap{j};
        obj.heap{j} = temp;
        temp = obj.priorities(i);
        obj.priorities(i) = obj.priorities(j);
        obj.priorities(j) = temp;
    end
    
    function p = parent(obj, k)
        % 返回堆中下标为k的元素的父节点下标
        p = floor(k / 2);
    end
    
    function c = leftChild(obj, k)
        % 返回堆中下标为k的元素的左子节点下标
        c = 2 * k;
    end
end