帮我用java语言解决基于优先队列式分支限界法解决装载问题，给出实例具体展示，给出具体实现代码

下面是基于优先队列式分支限界法解决装载问题的Java代码实现。

import java.util.*;

public class KnapsackProblem {
    static class Item implements Comparable<Item> {
        int weight;
        int value;
        double density;
        
        public Item(int weight, int value) {
            this.weight = weight;
            this.value = value;
            this.density = (double) value / weight;
        }
        
        @Override
        public int compareTo(Item other) {
            if (this.density > other.density) {
                return -1;
            } else if (this.density < other.density) {
                return 1;
            } else {
                return 0;
            }
        }
    }
    
    static class Node implements Comparable<Node> {
        int level;
        int profit;
        int weight;
        boolean[] include;
        
        public Node(int level, int profit, int weight, boolean[] include) {
            this.level = level;
            this.profit = profit;
            this.weight = weight;
            this.include = include.clone();
        }
        
        @Override
        public int compareTo(Node other) {
            return Integer.compare(other.profit, this.profit);
        }
    }
    
    public static int knapsack(int capacity, int[] weights, int[] values) {
        int n = weights.length;
        Item[] items = new Item[n];
        for (int i = 0; i < n; i++) {
            items[i] = new Item(weights[i], values[i]);
        }
        Arrays.sort(items);
        
        PriorityQueue<Node> queue = new PriorityQueue<>();
        boolean[] initialInclude = new boolean[n];
        queue.offer(new Node(-1, 0, 0, initialInclude));
        
        int maxProfit = 0;
        while (!queue.isEmpty()) {
            Node node = queue.poll();
            if (node.level == n - 1) {
                if (node.profit > maxProfit) {
                    maxProfit = node.profit;
                }
                continue;
            }
            int level = node.level + 1;
            boolean[] include = node.include.clone();
            include[level] = true;
            int weight = node.weight + items[level].weight;
            if (weight <= capacity) {
                int profit = node.profit + items[level].value;
                queue.offer(new Node(level, profit, weight, include));
            }
            int bound = bound(level, capacity, items, node);
            if (bound > maxProfit) {
                queue.offer(new Node(level, node.profit, node.weight, node.include.clone()));
            }
        }
        
        return maxProfit;
    }
    
    private static int bound(int level, int capacity, Item[] items, Node node) {
        int profit = node.profit;
        int weight = node.weight;
        for (int i = level + 1; i < items.length; i++) {
            if (weight + items[i].weight <= capacity) {
                weight += items[i].weight;
                profit += items[i].value;
            } else {
                profit += (capacity - weight) * items[i].density;
                break;
            }
        }
        return profit;
    }
    
    public static void main(String[] args) {
        int capacity = 50;
        int[] weights = {10, 20, 30};
        int[] values = {60, 100, 120};
        int maxProfit = knapsack(capacity, weights, values);
        System.out.println("Maximum profit is: " + maxProfit);
    }
}

在上面的代码中，我们定义了两个内部类`Item`和`Node`。`Item`表示物品，包含重量、价值和价值密度；`Node`表示搜索树中的节点，包含当前层数、当前总价值、当前总重量和是否选择了每个物品。

我们首先将物品按照价值密度从大到小排序，然后使用优先队列保存搜索树中的节点。每次从队列中取出价值最大的节点，根据它的状态进行分支。如果选择当前物品能够放入背包，则生成一个新节点，加入队列中。如果选择当前物品不能放入背包，则计算当前节点的上界，如果该上界大于当前最优解，则生成一个新节点，加入队列中。

最后，我们输出得到的最大价值即可。