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。 ```java 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。