二项队列是 堆序 的集合，也叫 森林。其中每一种形式都有约束。
二项树Bk由一个带有儿子的B0,B1,B2...组成，高度为k的二项树 恰好有2^k个结点。每一种高度只能出现一次...因此，只有1,2,4,8...等结点数目的二项树
deleteMin操作需要快速的找出跟的所有子树的能力，因此需要一般树的表示方法：
每个结点的儿子都在一个链表中，而且每个结点都有一个指向它的第一个儿子的指针。
二项树的每一个结点包括：数据，第一个儿子，以及右兄弟
下面是二项队列类构架及结点定义：
1 template <typename Comparable> 2 class BinomialQueue 3 { 4 public: 5 BinomialQueue(); 6 BinomialQueue(const Comparable & item); 7 BinomialQueue(const BinomialQueue & rhs); 8 ~BinomialQueue(); 9 10 bool isEmpty() const; 11 const Comparable & findMin() const; 12 13 void insert(const Comparable & x); 14 void deleteMin(); 15 void deleteMin(Comparable & minItem); 16 17 void makeEmpty(); 18 void merge(BinomialQueue & rhs); 19 20 const BinomialQueue & operator=(const BinomialQueue & rhs); 21 private: 22 struct BinomialNode 23 { 24 Comparable element; 25 BinomiaNode *leftChild; 26 BinomialNode *nextSibling; 27 28 BinomialNode(const Comparable & theElement,BinomialNode *lt,BinomialNode *rt) 29 :element(theElement),leftChild(lt),rightChild(rt) 30 { 31 } 32 }; 33 34 enum{DEFAULT_TREES = 1}; 35 36 int currentSize; 37 vector<BinomialNode *> const; 38 int capacity() const; 39 BinomialNode * combineTrees(BinomialNode *t1,BinomialNode *t2); 40 void makeEmpty(BinomialNode * & t); 41 BinomialNode * clone(BinomialNode *t) const; 42 };
合并同样大小的两棵树二项树的例程：
1 BinomialNode * combineTrees(BinomialNode *t1,BinomialNode *t2) 2 { 3 if(t2->element < t1->element) 4 return combineTrees(t2,t1); 5 t1->nextSibling = t1->leftChild; 6 t1->leftChild = t2; 7 return t1; 8 }
合并两个优先队列的例程：
1 void merge(BinomialQueue & rhs) 2 { 3 if(this==&rhs) 4 return; 5 currentSize += rhs.currentSize; 6 if(currentSize > capacity()) 7 { 8 int oldNumTrees = theTrees.size(); 9 int newNumTrees = max(theTrees.size(),rhs.theTrees.size())+1; 10 theTrees.resize(newNumTrees); 11 for(int i = oldNumTrees; i<newNumTrees; i++) 12 theTrees[i] = NULL; 13 } 14 BinomialNode *carry = NULL; 15 for(int i=0,j=1; j<=currentSize; i++,j*=2) 16 { 17 BinomialNode *t1 = theTrees[i]; 18 BinomialNode *t2 = i < rhs.theTrees.size()? rhs.theTrees[i]:NULL; 19 20 int whichCase = t1 == NULL ? 0 : 1; 21 whichCase += t2 == NULL? 0 : 2; 22 whichCase += carry == NULL? 0 : 4; 23 24 switch(whichCase) 25 { 26 case 0: 27 case 1: 28 break; 29 case 2: 30 theTrees[i] = t2; 31 rhs.theTrees[i] = NULL; 32 break; 33 case 4: 34 theTrees[i] = carry; 35 carry = NULL; 36 break; 37 case 3: 38 carry = combineTrees(t1,t2); 39 theTrees[i] = rhs.theTrees[i] = NULL; 40 break; 41 case 5: 42 carry = combineTrees(t1,carry); 43 theTrees[i] = NULL; 44 break; 45 case 6: 46 carry = combineTrees(t2,carry); 47 rhs.theTrees[i] = NULL; 48 break; 49 case 7: 50 theTrees[i] = carry; 51 carry = combineTrees(t1,t2); 52 rhs.theTrees[i] = NULL; 53 break; 54 } 55 } 56 for(int k = 0; k < rhs.theTrees.size(); k++) 57 rhs.theTrees[k] = NULL; 58 rhs.currentSize = 0; 59 }
deleteMin程序：
1 void deleteMin(Comparable & minItem) 2 { 3 if(isEmpty()) 4 throw UnderflowException(); 5 int minIndex = findMinIndex(); 6 minItem = theTrees[minIndex]->element; 7 8 BinomialNode *oldRoot = theTrees[minIndex]; 9 BinomialNode *deletedTree = oldRoot->leftChild; 10 delete oldRoot; 11 12 BinomialQueue deletedQueue; 13 deletedQueue.theTrees.resize(minIndex+1); 14 deletedQueue.currentSize = (1<<minIndex)-1; 15 for(int j = minIndex -1 ;j >= 0 ; j--) 16 { 17 deletedQueue.theTrees[ j ] = deletedTree; 18 deletedTree = deletedTree->nextSibling; 19 deletedQueue.theTrees[ j ]->nextSibling = NULL; 20 } 21 22 theTrees[minIndex] = NULL; 23 currentSize -= deletedQueue.currentSize + 1; 24 25 merge(deletedQueue); 26 } 27 int findMinIndex() const 28 { 29 int i; 30 int minIndex; 31 32 for( i = 0; theTrees[i] === NULL;i++) 33 ; 34 for(minIndex = i; i < theTrees.size();i++) 35 if(theTrees[i] != NULL && theTrees[i]->element < theTrees[minIndex]->element) 36 minIndex = i; 37 38 return minIndex; 39 }