二项队列 C++实现

二项队列是比左式堆与斜堆更好的选择。

二项队列不是一棵树，是由堆序的集合组成的，成为森林。堆序中每一颗树都有约束，都为二项树，每一个高度只有一个二项树，每层的高度由两个下一层高度的树组成。B1= B0 + B0。



二项队列的合并是相同高度的二项树合并，并且insert与deleteMin都是依靠合并这个操作实现的。

template<typename T>
class BinomialQueue
{
public:
BinomialQueue();
BinomialQueue(const T &t);
BinomialQueue(const BinomialQueue &rhs);
~BinomialQueue();

bool isEmpty() const;
const T & findMin() const;

void insert(const T &t);
void deleteMin();
void deleteMin(T & t);

void makeEmpty();
void merge(BinomialQueue & rhs);

const BinomialQueue & operator= (const BinomialQueue &rhs);

private:
struct BinomialNode
{
T element;
BinomialNode * leftChild;
BinomialNode * nextSibling;

BinomialNode(const T & t, BinomialNode * lc, BinomialNode *ns) : element(t), leftChild(lc), nextSibling(ns) {}

};
enum { DEFAULT_TREES = 1};
int currentSize;
vector<BinomialNode *> theTrees;

int findMinIndex() const;
int capacity() const;
BinomialNode * combineTrees(BinomialNode * t1, BinomialNode * t2);
void makeEmpty(BinomialNode * & t);
BinomialNode * clone(BinomialNode * t) const;
};

template<typename T>
BinomialNode * BinomialQueue<T>::combineTrees(BinomialNode * t1, BinomialNode * t2)
{
if (t2->element < t1->element)
return combineTrees(t2, t1);
t2->nextSibling = t1->leftChild;
t1->leftChild = t2;
return t1;
}

template<typename T>
void BinomialQueue<T>::merge(BinomialQueue & rhs)
{
if (this == &rhs)
return;
currentSize == rhs.currentSize > theTrees.size() ? rhs.currentSize + 1 : theTrees.size() + 1;
//上一层树合并生成的下一层树，没有则为空
BinomialNode * carry = nullptr;
//书上写的这个j是什么意思啊？ 求解
for (int i = 0, j = 1; i < currentSize; ++i, j *= 2)
{
BinomialNode * t1 = theTrees[i];
BinomialNode * t2 = i < rhs.theTrees.size() ? rhs.theTrees[i] : nullptr;
int whichCase = t1 == nullptr ? 0 : 1;
whichCase += t2 == nullptr ? 0 : 2;
whichCase += carry == nullptr ? 0 : 4;
switch (whichCase)
{
case 0://没有树
case 1://就有本身这棵树
break;
case 2://只有rhs有
theTrees[2] = t2;
rhs.theTrees[2] = nullptr;
break;
case 4://只有carry
theTrees[i] = carry;
carry = nullptr;
break;
case 3://有本身和rhs
carry = combineTrees(t1, t2);
theTrees[i] = rhs.theTrees[i] = nullptr;
break;
case 5://本身和carry
carry = combineTrees(t1, carry);
theTrees[i] = nullptr;
break;
case 6://rhs与carry
carry = combineTrees(t2, carry);
rhs.theTrees[i] = nullptr;
break;
case 7://本身、rhs、carry都有
theTrees[i] = carry;
carry = combineTrees(t1, t2);
rhs.theTrees[i] = nullptr;
break;
}
}
for (int k = 0; k < rhs.theTrees.size(); ++k)
rhs.theTrees[k] = nullptr;
rhs.currentSize = 0;
}