二进制搜索树最佳拟合算法:输出不正确

Binary Search Tree Best Fit Alg.: Incorrect Output

本文关键字：输出不正确算法拟合搜索树最佳二进制更新时间：2023-10-16

免责声明：这是学校作业

大家好。我已经在这个Bin Packing程序上辛苦工作了两周，我还有最后一个障碍要克服：我的二进制搜索树的查找功能给了我不正确的结果。

二进制搜索树.cpp

#include "BinarySearchTree.h"
void BinarySearchTree::insert(int capacity, int binNumber)
{
    // Insert the Pair into the tree. Overwrite existing
    // pair, if any, with same key.
    // find place to insert
    BinaryTreeNode *p = root,
                         *pp = NULL;
    while (p != NULL)
    {// examine p->capacity
        pp = p;
        // move p to a child
        if (capacity <= p->capacity)
            p = p->leftChild;
        else
            p = p->rightChild;
    }
    // get a node for the Pair and attach to pp
    BinaryTreeNode *newNode = new BinaryTreeNode (capacity, binNumber);
    if (root != NULL) // the tree is not empty
        if (capacity <= pp->capacity)
            pp->leftChild = newNode;
        else
            pp->rightChild = newNode;
    else
        root = newNode; // insertion into empty tree
    treeSize++;
}
void BinarySearchTree::erase(BinaryTreeNode *n)
{
    // Delete the pair, if any, whose key equals n.
    // search for node with key theKey
    BinaryTreeNode *p = root,
                         *pp = NULL;
    while (p != NULL && p->capacity != n->capacity)
    {// move to a child of p
        pp = p;
        if (n->capacity < p->capacity)
            p = p->leftChild;
        else
            p = p->rightChild;
    }
    if (p == NULL)
        return; // no pair with key theKey
    // restructure tree
    // handle case when p has two children
    if (p->leftChild != NULL && p->rightChild != NULL)
    {// two children
        // convert to zero or one child case
        // find largest element in left subtree of p
        BinaryTreeNode *s = p->leftChild,
                *ps = p;  // parent of s
        while (s->rightChild != NULL)
        {// move to larger element
            ps = s;
            s = s->rightChild;
        }
        // move largest from s to p, can't do a simple move
        // p->capacity= s->capacity as key is const
        BinaryTreeNode *q = new BinaryTreeNode (s->capacity,s->binNumber, p->leftChild, p->rightChild, p->parent);
        if (pp == NULL)
            root = q;
        else if (p == pp->leftChild)
            pp->leftChild = q;
        else
            pp->rightChild = q;
        if (ps == p) pp = q;
        else pp = ps;
        delete p;
        p = s;
    }
    // p has at most one child
    // save child pointer in c
    BinaryTreeNode *c;
    if (p->leftChild != NULL)
        c = p->leftChild;
    else
        c = p->rightChild;
    // delete p
    if (p == root)
        root = c;
    else
    {// is p left or right child of pp?
        if (p == pp->leftChild)
            pp->leftChild = c;
        else pp->rightChild = c;
    }
    treeSize--;
    delete p;
}
BinaryTreeNode* BinarySearchTree::find(const int objectSize) const
{
    // Return pointer to pair with smallest key >= objectSize.
    // Return NULL if no element has key >= objectSize.
    BinaryTreeNode *currentNode = root,
                   *bestElement = NULL; // element with smallest key
                                     // >= theKey found so far
    // search the tree
    while (currentNode != NULL) {
        // is currentNode->capacity a candidate?
        if (currentNode->capacity >= objectSize)
        {
            // smaller keys in left subtree only
            bestElement = currentNode;
            currentNode = currentNode->leftChild;
        }
        else if (currentNode->capacity < objectSize)
        {
            // no, currentNode->capacity is too small
            // try right subtree
            currentNode = currentNode->rightChild;
        }
    }
    return bestElement;
}

BinaryTreeNode.h

struct BinaryTreeNode
{
    public:
        BinaryTreeNode *leftChild;
        BinaryTreeNode *rightChild;
        BinaryTreeNode *parent;
        int capacity;
        int binNumber;
        BinaryTreeNode() {leftChild = rightChild = parent = NULL;}
        BinaryTreeNode(const int& c, const int& b):capacity(c), binNumber(b)
        {
            leftChild = rightChild = parent = NULL;
        }
        BinaryTreeNode(const int& c, const int& b, BinaryTreeNode* l, BinaryTreeNode* r, BinaryTreeNode* p):capacity(c), binNumber(b)
        {
            leftChild = l;
            rightChild = r;
            parent = p;
        }
};

BinPacking.cpp

void BinPacking::bestFitPack(int *objectSize, int numberOfObjects, int binCapacity)
{// Output best-fit packing into bins of size binCapacity.
 // objectSize[1:numberOfObjects] are the object sizes.
   int n = numberOfObjects;
   int binsUsed = 0;
   BinarySearchTree theTree;  // tree of bin capacities
   BinaryTreeNode theBin;
   // pack objects one by one
   for (int i = 1; i <= n; i++)
   {// pack object i
      // find best bin
       BinaryTreeNode *bestBin = theTree.find(objectSize[i]);
      if (bestBin == NULL)
      {// no bin large enough, start a new bin
         theBin.capacity = binCapacity;
         theBin.binNumber = ++binsUsed;
      }
      else
      {// remove best bin from theTree
         theBin = *bestBin;
         theTree.erase(bestBin);
      }
      cout << "Pack object " << i << " in bin " << theBin.binNumber << endl;
      // insert bin in tree unless bin is full
      theBin.capacity -= objectSize[i];
      if (theBin.capacity > 0)
         theTree.insert(theBin.capacity, theBin.binNumber);
   }
}

用户输入到主（未显示）

# of objects = 12
Bin capacity = 6
Sizes of objects:
object 1  = 2 
object 2  = 5
object 3  = 5
object 4  = 1
object 5  = 1
object 6  = 3
object 7  = 4
object 8  = 6
object 9  = 2
object 10 = 5
object 11 = 6
object 12 = 1

预期输出

Pack object 1 in bin 1 
Pack object 2 in bin 2 
Pack object 3 in bin 3 
Pack object 4 in bin 2 
Pack object 5 in bin 3 
Pack object 6 in bin 1 
Pack object 7 in bin 4 
Pack object 8 in bin 5 
Pack object 9 in bin 4 
Pack object 10 in bin 6 
Pack object 11 in bin 7 
Pack object 12 in bin 1

电流输出

Pack object 1 in bin 1 
Pack object 2 in bin 2 
Pack object 3 in bin 3 
Pack object 4 in bin 3 
Pack object 5 in bin 3 
Pack object 6 in bin 1 
Pack object 7 in bin 4 
Pack object 8 in bin 5 
Pack object 9 in bin 4 
Pack object 10 in bin 6 
Pack object 11 in bin 7 
Pack object 12 in bin 6

我知道我快要完成这项任务了。我知道问题出在哪里，但我一直没能解决。求你了，你能帮我吗？

错误出现在代码的erase():中

while (p != NULL && p->capacity != n->capacity)
{// move to a child of p
    pp = p;
    if (n->capacity < p->capacity)
        p = p->leftChild;
    else
        p = p->rightChild;
}

因为您要传入一个特定的节点来擦除，因为树中的多个bin可以具有相同的当前剩余容量，并且因为您的"最佳bin"逻辑返回递归中的最后一个最佳匹配，你的erase()函数可能会erase()错误的节点，事实上很可能会。（如果你的匹配代码总是返回第一个"最佳匹配"，我不确定你会有问题，尽管代码很脆弱。）

如果同等容量的垃圾箱真的无法区分，这实际上就不是问题了。然而，代码中的每个bin都有一个唯一的binNumber，因此您会在您所说的分配到的bin和实际擦除和重新插入的bin之间断开连接。

由于您要传递一个指向树中某个元素的指针，因此您应该直接在这里进行指针比较。

while (p != n)
{// move to a child of p
    pp = p;
    if (p->capacity >= N->capacity)
        p = p->leftChild;
    else
        p = p->rightChild;
}

删除不在树中的节点无论如何都是非法的，因此唯一有效的终止条件是找到该节点。

一旦你找到了要擦除的节点，下面的伪代码应该正确地旋转树：

if (pp->capacity >= n->capacity)
    parent_to_me = &pp->leftChild;
else
    parent_to_me = &pp->rightChild;
if (!node->leftChild && !node->rightChild)
{
    // point to nothing
    *parent_to_me = NULL;
} else if (node->leftChild)
{
    if (!node->rightChild)
    {
        // left child, but no right child:
        // promote left child to self
        *parent_to_me = node->leftChild;
    } else
    {
        // left and right children:  Make right child the
        // right child of the right-most left sub-child.
        // Replace myself with left child.
        Node *rmlsc = node->leftChild;
        while (rmlsc->rightChild)
            rmlsc = rmlsc->rightChild;
        rmlsc->rightChild = node->rightChild;
        *parent_to_me = node->leftChild;
    }
} else
    *parent_to_me = node->rightChild;