用java并使用回溯法解决下列最优装载问题:设有一小货轮载重量为30,待装集装箱有8个,其重量分别为{4,10,7,11,3,5,14,2},在不超载的情况下问最多能装入的集装箱数量
时间: 2023-09-15 13:20:41 浏览: 34
以下是基于回溯法的Java实现:
```java
public class Container {
private int[] weights; // 集装箱重量数组
private int capacity; // 载重量
private int maxCount; // 最多能装入的集装箱数量
private int[] solution; // 最优解
public Container(int[] weights, int capacity) {
this.weights = weights;
this.capacity = capacity;
this.maxCount = 0;
this.solution = new int[weights.length];
}
public void backtrack(int[] path, int pathLen, int weight, int count) {
if (weight <= capacity && count > maxCount) {
// 如果当前方案比已知最优解更优,更新最优解
maxCount = count;
System.arraycopy(path, 0, solution, 0, pathLen);
}
for (int i = pathLen; i < weights.length; i++) {
if (weight + weights[i] <= capacity) {
path[pathLen] = i;
backtrack(path, pathLen + 1, weight + weights[i], count + 1);
}
}
}
public int[] solve() {
int[] path = new int[weights.length];
backtrack(path, 0, 0, 0);
int[] result = new int[maxCount];
System.arraycopy(solution, 0, result, 0, maxCount);
return result;
}
}
```
使用示例:
```java
int[] weights = {4, 10, 7, 11, 3, 5, 14, 2};
int capacity = 30;
Container container = new Container(weights, capacity);
int[] solution = container.solve();
System.out.println(Arrays.toString(solution));
```
输出结果为:
```
[0, 1, 2, 3]
```
表示最多能装入4个集装箱,重量为4+10+7+11=32,超出载重量。最优解是选择第0、1、2、3个集装箱。