Question

This algorithm对于我的基本编程技能是如此先进，我只是看不出如何实现它。我在一个新问题中发布这个问题，因为我不能继续打扰那个在上一个问题的评论部分中单独给我算法的人。

MaxSet(node) = 1 if "node" is a leaf
MaxSet(node) = Max(1 + Sum{ i=0..3: MaxSet(node.Grandchildren[i]) },  
                       Sum{ i=0..1: MaxSet(node.Children[i])      })

非常感谢mehrdad算法。

这里的问题是实现两条求和线的一部分，我该怎么做？我需要标记此算法选择的每个节点。它只是节点类中设置为true的“标记”变量。我不明白它是否决定选择一个节点？

编辑到目前为止包含我的代码：

public int maxSet(Posisjon<E> bt){
        if (isExternal(bt)){
            return 1; 
        }
        return Math.max(1 + helper1(bt), helper2(bt));
    }

private int helper1(Posisjon<E> node){
    int tmp = 0; 
    if (hasLeft(node)){
        if(hasLeft((Position<E>)node.leftChild())){
            tmp += maxSet(node.leftChild().leftChild());
        }
        if(hasRight((Position<E>)node.leftChild())){
            tmp += maxSet(node.leftChild().rightChild());
        }
    }
    if(hasRight(node)){
        if(hasLeft((Position<E>)node.rightChild())){
            tmp += maxSet(node.leftChild().leftChild());
        }
        if(hasRight((Position<E>)node.rightChild())){
            tmp += maxSet(node.leftChild().rightChild());
        }
    }
    return tmp; 
}
private int helper2(Posisjon<E> node){
    int tmp = 0; 
    if(hasLeft(node)){
        tmp +=maxSet(node.leftChild());
    }
    if(hasRight(node)){
        tmp +=maxSet(node.rightChild());
    }
    return tmp; 
}

这似乎有效，现在还剩下什么。是否实际将节点标记为已选择？我会这样做吗？

更新了代码：

public ArrayList<Posisjon<E>> getSelectionSet(Posisjon<E> bt, ArrayList<Posisjon<E>> s){
        if(bt.marked){
            s.add(bt);
        }
        if(hasLeft(bt)){
            if(hasLeft(bt.leftChild())){
                getSelectionSet(bt.leftChild().leftChild(),s);
            }
            if(hasRight(bt.leftChild())){
                getSelectionSet(bt.leftChild().rightChild(),s);
            }
        }
        if(hasRight(bt)){
            if(hasLeft(bt.rightChild())){
                getSelectionSet(bt.rightChild().leftChild(),s);
            }
            if(hasRight(bt.rightChild())){
                getSelectionSet(bt.rightChild().rightChild(),s);
            }
        }
        return s; 
    }

public int maxSet(Posisjon<E> bt){
        if (bt.visited){
            return bt.computedMax; 
        }
        bt.visited = true; 
        int maxIfCurrentNodeIsSelected = 1 + helper1(bt);
        int maxIfCurrentNodeIsNotSelected = helper2(bt);
        if (maxIfCurrentNodeIsSelected > maxIfCurrentNodeIsNotSelected){
            bt.marked = true; 
            bt.computedMax = maxIfCurrentNodeIsSelected; 
        }else{
            bt.marked = false; 
            bt.computedMax = maxIfCurrentNodeIsNotSelected; 
        }
        return maxSet(bt);
    }

提交后，我会发布完整的代码！

Answer 1

您目前每次都没有记忆该函数的返回值。每次拨打maxSet时，都应检查是否已经计算了结果。如果有，请退回。如果您还没有计算它并将其存储在某个地方。否则，您的算法效率低下。（这种方法称为“动态编程”。了解它。）

// pseudocode:
public int maxSet(Posisjon<E> bt){
    if (visited[bt])
        return computedMax[bt];

    visited[bt] = true;        

    // You don't need to manually check for being a leaf
    // For leaves 'maxIfCurrentNodeIsSelected' is always larger.
    int maxIfCurrentNodeIsSelected = 1 + helper1(bt);
    int maxIfCurrentNodeIsNotSelected = helper2(bt);

    if (maxIfCurrentNodeIsSelected > maxIfCurrentNodeIsNotSelected) {
         shouldSelect[bt] = true;
         computedMax[bt] = maxIfCurrentNodeIsSelected;
    } else {
         shouldSelect[bt] = false;
         computedMax[bt] = maxIfCurrentNodeIsNotSelected;
    }
}

public Set getSelectionSet(Posisjon<E> bt, Set s) {
    if (shouldSelect[bt]) {
        s.Add(bt);

        // You should check for nulls, of course
        getSelectionSet(bt.leftChild.leftChild, s);
        getSelectionSet(bt.leftChild.rightChild, s);
        getSelectionSet(bt.rightChild.leftChild, s);
        getSelectionSet(bt.rightChild.rightChild, s);
    } else {
        getSelectionSet(bt.leftChild, s);
        getSelectionSet(bt.rightChild, s);
    }
    return s;
}

在您调用getSelectionSet后，

使用根节点调用Set并使用空maxSet作为参数。

Java需要帮助实现算法

1 个答案: