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Binomial Queue

:material-circle-edit-outline: 约 157 个字 :fontawesome-solid-code: 84 行代码 :material-image-multiple-outline: 4 张图片 :material-clock-time-two-outline: 预计阅读时间 3 分钟

Structure

  • Definition:

    A binomial queue is not a heap-ordered tree, but rather a collection of heap-ordered trees, known as a forest. Each heap-ordered tree is a binomial tree.
    A priority queue of any size can be uniquely represented by a collection of binomial trees.

Example

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Operations

  • Findmin:

    The minimum key is in one of the roots. There are at most \(\lceil log_2N \rceil\) roots, hence the time complexity is \(O(logN)\).(Can be stored and updated in constant time)

  • Merge:

    Must keep the trees in the binomial queue sorted by height. \(T_p=O(logN)\)

  • Insert:

    If the smallest nonexistent binomial tree is Bi , then \(T_p=Const·(i+1)\). Performing N Inserts on an initially empty binomial queue will take O(N) worst-case time. Hence the average time is constant.

  • DeleteMin:

illustration

alt text

Implementation

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   typedef struct BinNode *Position;
   typedef struct Collection *BinQueue;
   typedef struct BinNode *BinTree;  /* missing from p.176 */

   struct BinNode 
   { 
    ElementType     Element;
    Position        LeftChild;
    Position        NextSibling;
   } ;

   struct Collection 
   { 
    int         CurrentSize;  /* total number of nodes */
    BinTree TheTrees[ MaxTrees ];
   } ;
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   BinTree
   CombineTrees( BinTree T1, BinTree T2 )
   {  /* merge equal-sized T1 and T2 */
    if ( T1->Element > T2->Element )
        /* attach the larger one to the smaller one */
        return CombineTrees( T2, T1 );
    /* insert T2 to the front of the children list of T1 */
    T2->NextSibling = T1->LeftChild;
    T1->LeftChild = T2;
    return T1;
   }
   BinQueue  Merge( BinQueue H1, BinQueue H2 )
   {    BinTree T1, T2, Carry = NULL;   
    int i, j;
    if ( H1->CurrentSize + H2-> CurrentSize > Capacity )  ErrorMessage();
    H1->CurrentSize += H2-> CurrentSize;
    for ( i=0, j=1; j<= H1->CurrentSize; i++, j*=2 ) {
        T1 = H1->TheTrees[i]; T2 = H2->TheTrees[i]; /*current trees */
        switch( 4*!!Carry + 2*!!T2 + !!T1 ) { 
        case 0: /* 000 */
        case 1: /* 001 */  break;   
        case 2: /* 010 */  H1->TheTrees[i] = T2; H2->TheTrees[i] = NULL; break;
        case 4: /* 100 */  H1->TheTrees[i] = Carry; Carry = NULL; break;
        case 3: /* 011 */  Carry = CombineTrees( T1, T2 );
                        H1->TheTrees[i] = H2->TheTrees[i] = NULL; break;
        case 5: /* 101 */  Carry = CombineTrees( T1, Carry );
                        H1->TheTrees[i] = NULL; break;
        case 6: /* 110 */  Carry = CombineTrees( T2, Carry );
                        H2->TheTrees[i] = NULL; break;
        case 7: /* 111 */  H1->TheTrees[i] = Carry; 
                        Carry = CombineTrees( T1, T2 ); 
                        H2->TheTrees[i] = NULL; break;
        } /* end switch */
    } /* end for-loop */
    return H1;
   }
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   ElementType  DeleteMin( BinQueue H )
   {    BinQueue DeletedQueue; 
    Position DeletedTree, OldRoot;
    ElementType MinItem = Infinity;  /* the minimum item to be returned */  
    int i, j, MinTree; /* MinTree is the index of the tree with the minimum item */

    if ( IsEmpty( H ) )  {  PrintErrorMessage();  return –Infinity; }

    for ( i = 0; i < MaxTrees; i++) {  /* Step 1: find the minimum item */
        if( H->TheTrees[i] && H->TheTrees[i]->Element < MinItem ) { 
        MinItem = H->TheTrees[i]->Element;  MinTree = i;    } /* end if */
    } /* end for-i-loop */
    DeletedTree = H->TheTrees[ MinTree ];  
    H->TheTrees[ MinTree ] = NULL;   /* Step 2: remove the MinTree from H => H’ */ 
    OldRoot = DeletedTree;   /* Step 3.1: remove the root */ 
    DeletedTree = DeletedTree->LeftChild;   free(OldRoot);
    DeletedQueue = Initialize();   /* Step 3.2: create H” */ 
    DeletedQueue->CurrentSize = ( 1<<MinTree ) – 1;  /* 2MinTree – 1 */
    for ( j = MinTree – 1; j >= 0; j – – ) {  
        DeletedQueue->TheTrees[j] = DeletedTree;
        DeletedTree = DeletedTree->NextSibling;
        DeletedQueue->TheTrees[j]->NextSibling = NULL;
    } /* end for-j-loop */
    H->CurrentSize  – = DeletedQueue->CurrentSize + 1;
    H = Merge( H, DeletedQueue ); /* Step 4: merge H’ and H” */ 
    return MinItem;
   }

Proof

  • Claim: A binomial queue of N elements can be built by N successive insertions in O(N) time.

Proof

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