Bzoj_3224_ General Balance tree (TREAP)

Describe

Treap template questions. The following actions are supported:

1. Insert;

2. Deletion;

3.rank (x);

4.kth (k);

5.pre (x);

6.SUC (x);

Analysis

When you write more, you'll be ripe ... Yesterday read the book, debugging today when knocked many times, feeling full of tree+heap.

P.S.

The 1.RANK,KTH,PRE,SUC function can be written in a non-recursive form.

The 2.remove function is written in two ways:

(1). LRJ petition on the method:

If the node you want to delete has only one subtree, use that subtree instead of the node you want to delete, and delete the node you want to delete. If there are two subtrees, rotate the Keng Gen subtree with a small priority value, and then recursively delete the node to be deleted in another subtree, each time this rotation will reduce the tree, and eventually the first case.

(2). The method that appears in the solution:

If the node you want to delete has no subtree, you can delete it directly. If there are subtrees, either two or one, rotate the subtree with a small precedence value (note that null precedence is set to INF), and then recursively delete nodes in another subtree, and each of these rotations will reduce the tree. The first situation will eventually be achieved.

#include <cstdio>#include<cstdlib>using namespacestd;Const intoo=~0u<<1;structtreap{structnode{Node* ch[2]; intV,s,r,c; Node (intV,node *t): V (v) {ch[0]=ch[1]=t; R=rand (); S=c=1; } BOOL operator< (ConstNode &AMP;RHS)Const{returnr<RHS.R;} voidPush_up () {s=ch[0]->s+ch[1]->s+C;} }*root,*NULL; Treap () {NULL=NewNode0,0); NULL->s=NULL->c=0; NULL->r=Oo; Root=NULL; }    voidRotate (node* &o,BOOLd) {Node* k=o->ch[!d]; o->ch[!d]=k->ch[d]; k->ch[d]=N; o->push_up (); K->push_up (); o=K; }    voidInsert (node* &o,intx) {        if(o==NULL) o=NewNode (x,NULL); Else{            if(o->v==x) {o->c++; o->s++; }            Else{                BOOLD=x>o->v; Insert (o-ch[d],x); if(o->ch[d]<o) Rotate (o,!d); o-push_up (); }        }    }    voidRemove (node* &o,intx) {        if(o->v==x) {            if(o->c>1) o->c--; Else{                if(o->ch[0]!=NULL&&o->ch[1]!=NULL){                    BOOLd=o->ch[0]<o->ch[1]; Rotate (o,d); Remove (o-ch[d],x); }                Else{node* u=o; if(o->ch[0]==NULL) o=o->ch[1]; Elseo=o->ch[0]; Deleteu; }            }        }        Else{            BOOLD=x>o->v; Remove (o-ch[d],x); }        if(o!=NULL) o->push_up (); }    intKTH (node* o,intk) {        ints=o->ch[0]->s+o->C; if(k>o->ch[0]->s&&k<=s)returnO->W; if(k<=o->ch[0]->s)returnKTH (o->ch[0],k); Else returnKTH (o->ch[1],k-s); }    intRank (node *o,intx) {        ints=o->ch[0]->s+o->B; if(X==O-&GT;V)returno->ch[0]->s+1; if(X&LT;O-&GT;V)returnRank (o->ch[0],x); Else returnS+rank (o->ch[1],x); }    intPreintx) {Node* t=Root; intret=0;  while(t!=NULL){            if(t->v<x) {ret=t->v; T=t->ch[1]; }            Elset=t->ch[0]; }        returnret; }    intSucintx) {Node*t=Root; intret=0;  while(t!=NULL){            if(t->v>x) {ret=t->v; T=t->ch[0]; }            Elset=t->ch[1]; }        returnret; }}tree;intMain () {intn,a,b; scanf ("%d",&N);  while(n--) {scanf ("%d%d",&a,&b); Switch(a) { Case(1): Tree.insert (TREE.ROOT,B); Break;  Case(2): Tree.remove (TREE.ROOT,B); Break;  Case(3):p rintf ("%d\n", Tree.rank (tree.root,b)); Break;  Case(4):p rintf ("%d\n", tree.kth (tree.root,b)); Break;  Case(5):p rintf ("%d\n", Tree.pre (b)); Break;  Case(6):p rintf ("%d\n", TREE.SUC (b)); Break; }    }    return 0;}

Bzoj_3224_ General Balance tree (TREAP)