Question

我读到的是:Binomial queue 行动here。

联系的底部是指

B. 血清

deletemin operation requires ability to find all subtrees of the root. Thus children of each node should be available (say a linked list)
deletemin requires that the children be ordered by the size of their subtrees.
we need to make sure it is easy to merge tress. Two binomial trees can be merged only if they have the same size, hence the size of the tree must be stored in the root. While merging, one of the trees becomes the last child of the other, so we should keep track of the last child of each node. A good data structure to use is a circular doubly linked list what each node is of the following form:

data | first |left    | right |rank No. of |
--------------------------------------------
       child |sibling |sibling| children

以上是作者所称的“No. Of? 如果有的话,请作解释。

Answer 1

据我所知,他试图说:我们储存了<条码>>>rank<>/条码>,该编码与《儿童第<条码>有误。 (通常如何界定这些树木的等级)。因此,你只是在每个节点储存:

data represents the element in the tree
first represents a pointer to the linked list of children (i.e. a pointer to the first child)
left is a pointer to the left neighbour
right is a pointer to the right neighbour
rank is simply the rank of the binomial tree

Answer 2

说明要求 “只有具有相同规模的 bin树才能合并,因此树木的规模必须储存在根基中”。

看来,提交人不是“小林区”而是“儿童人数”领域。这令人困惑,但对于执行工作来说,这是很不错的,因为小树的面积是2 ># of children}。因此,你可以比较儿童人数,而不是比较小林面积。