二项队列

Posted 2020-09-30 啊哈咧

堆序树的集合，森林

二项树 Bk 由一个带有儿子 B1 B2 ... Bk-1 的根组成 ，高度为 k 的二项树恰好有 2^k 个节点

目标：用二项树的集合惟一表示任意大小的优先队列

树的合并只在乎有没有保持堆序性，兄弟节点间没有要求

合并 2 个优先队列部分代码

T1=H1->TheTrees[i];
T2=H2->TheTrees[i];
switch (!!T1 + 2*!!T2 + 4*!!Carry) {
    case 0:
    case 1:
        break;
    case 2:
        H1->TheTrees[i] = T2;
        H2->TheTrees[i] = NULL;
        break;
    case 4:
        H1->TheTrees[i] = Carry;
        Carry = NULL;
        break;
    case 3:
        carry = CombineTrees(T1, T2);
        H1->TheTrees[i] = H2->TheTrees[i] = NULL;
        break;
    case 5:
        Carry = CombineTrees(T1, Carry);
        H1->TheTrees[i] = NULL;
        break;
    case 6:
        Carry = CombineTrees(T2, Carry);
        H2->TheTrees[i] = NULL;
        break;
    case 7:
        H1->TheTrees[i] = Carry;
        Carry = CombineTrees(T1, T2);
        H2->TheTrees[i] = NULL;
        break;
}

 CombineTrees代码如下

BinTree CombineTrees(BinTree T1, BinTree T2) {
    if (T1->Element > T2->Element) {
        return CombineTrees(T2, T1);
    }
    T2->NextSibling = T1->LeftChild;
    T1->LeftChild = T2;
    return T1;
}