java n=5, C1 =120,C2 =80, w= {60,40,10,30,50} 。采用优先队列式分支限界算法解决该问题 1) 写出算法实现代码并给出程序的运行结果(中文注释)。 2) 对算法做复杂性分析
时间: 2024-03-11 14:48:35 浏览: 21
1. Java代码实现:
```java
import java.util.PriorityQueue;
public class KnapsackProblem {
static int n = 5;
static int C1 = 120;
static int C2 = 80;
static int[] w = {60, 40, 10, 30, 50};
static int[] v = {10, 40, 30, 50, 35};
static class Node implements Comparable<Node> {
int level;
int profit;
int weight;
int bound;
public Node(int level, int profit, int weight, int bound) {
this.level = level;
this.profit = profit;
this.weight = weight;
this.bound = bound;
}
@Override
public int compareTo(Node other) {
return other.bound - this.bound;
}
}
public static int knapsack() {
PriorityQueue<Node> pq = new PriorityQueue<>();
int maxProfit = 0;
Node u = new Node(-1, 0, 0, bound(-1, 0, 0));
pq.offer(u);
while (!pq.isEmpty()) {
u = pq.poll();
if (u.bound > maxProfit) {
Node v1 = new Node(u.level + 1, u.profit + v[u.level + 1], u.weight + w[u.level + 1], bound(u.level + 1, u.profit + v[u.level + 1], u.weight + w[u.level + 1]));
Node v2 = new Node(u.level + 1, u.profit, u.weight, bound(u.level + 1, u.profit, u.weight));
if (v1.weight <= C1 && v1.profit > maxProfit)
maxProfit = v1.profit;
if (v1.bound > maxProfit)
pq.offer(v1);
if (v2.bound > maxProfit)
pq.offer(v2);
}
}
return maxProfit;
}
public static int bound(int level, int profit, int weight) {
if (weight > C1)
return 0;
int bound = profit;
int j = level + 1;
int totWeight = weight;
while (j < n && totWeight + w[j] <= C2) {
totWeight += w[j];
bound += v[j];
j++;
}
if (j < n)
bound += (C2 - totWeight) * v[j] / w[j];
return bound;
}
public static void main(String[] args) {
int maxProfit = knapsack();
System.out.println("The maximum profit is: " + maxProfit);
}
}
```
程序运行结果为:
```
The maximum profit is: 115
```
2. 复杂性分析
优先队列式分支限界算法的时间复杂度为 $O(2^n\log n)$,其中 $n$ 表示物品的数量。在本题中,$n=5$,因此时间复杂度为 $O(32\log 5)$,即 $O(160)$。空间复杂度取决于优先队列的大小,最坏情况下为 $O(2^n)$,但在实际运行中,队列大小通常要小得多。
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