几种平衡树

#include<cstdio>
#include<iostream>
#include<cstring>
#define ls(x) (x->ch[0])
#define rs(x) (x->ch[1])
#define sz(x) (x->size)
#define p(x) (x->p)
using namespace std;
const int INF=0x7fffffff;
struct son
{
    int v,size;
    son *ch[2],*p;
    son(int x)
    {
        v=x;size=1;
    }
};
son *root,*null;
son* newson(int val)
{
    son *x=new son(val);
    ls(x)=rs(x)=p(x)=null;
    return x;
}
void pushup(son *x){sz(x)=sz(ls(x))+sz(rs(x))+1;}
int rank(int val)
{
    son *x=root;
    int ans=0;
    while(x!=null)
    {
        if(val>x->v){ans+=(sz(ls(x))+1);x=rs(x);}
        else x=ls(x);
    }
    return ans;
}
son* kth(int k)
{
    //printf("sha bi qin shi yu
");
    son *x=root;
    while(x!=null)
    {
        if(sz(ls(x))+1==k)return x;
        if(sz(ls(x))+1>k)x=ls(x);
        else {k-=(sz(ls(x))+1);x=rs(x);}
    }
    return null;
}
int islc(son *x){return ls(p(x))==x;}
// is left child ???
//x在右，就左旋  在左，就右旋 
void rot(son *x,int d)
{
    son *y=x->ch[d^1];
    if(p(x)!=null)p(x)->ch[islc(x)^1]=y;//祖父儿子=y
    else root=y;
    p(y)=p(x);//把祖父儿子设成y 
    x->ch[d^1]=y->ch[d];//
    if(y->ch[d])p(y->ch[d])=x;//把y的d同向儿子改成x的d逆向儿子 
    y->ch[d]=x;p(x)=y;//把x的爸爸改成y 
    pushup(x);pushup(y); 
}
void splay(son *x,son *t=null)//把x旋到以t为祖父节点 
{
    for(son *rt=p(x);rt!=t;rt=p(x))
    {
        if(p(rt)==t)
        {
            rot(rt,islc(x));
            return ;
        }
        if(islc(x)==islc(rt))rot(p(rt),islc(rt));//一字型旋转 第一步 
        else rot(rt,islc(x));//之字形旋转 第一步 
        rot(p(x),islc(x));//第二步 (此时x已经转上去)
    }
}
void add(int val)
{
    int pos=rank(val);//val排名-1 
    splay(kth(pos));//把前一个转到root
    splay(kth(pos+1),root);//把pos+1的转到root右儿子
    //val 卡在pos和pos+1之间 
    son *x=newson(val);
    root->ch[1]->ch[0]=x;
    p(x)=root->ch[1];
    pushup(root->ch[1]);pushup(root);
}
void del(int val)
{
    int pos=rank(val)+1;
    splay(kth(pos-1));splay(kth(pos+1),root);
    //把val卡在pos-1和pos+1之间
    root->ch[1]->ch[0]=null;
    pushup(root->ch[1]);pushup(root);
}
int n;

int main(){
    //freopen("phs.in","r",stdin);
    //freopen("phs.out","w",stdout);
    null=new son(0);sz(null)=0;
    root=newson(-INF);
    root->ch[1]=newson(INF);//即把一开始的区间调成-INF到INF
    p(root->ch[1])=root; 
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
    {
        int opt,x;scanf("%d%d",&opt,&x);
        switch(opt)
        {
            case 1:add(x);break;
            case 2:del(x);break;
            case 3:printf("%d
",rank(x));break;
            //返回值就是rank的原因：********
            //一开始root的值=-INF root右儿子=INF *******
            //每次查找一定会到这两个节点，而他们size=1 
            case 4:printf("%d
",kth(x+1)->v);break;
            //不能轻易在rank上+1，必须在x上+1 
            case 5:printf("%d
",kth(rank(x))->v);break;
            case 6:printf("%d
",kth(rank(x+1)+1)->v);break;
        }
    }
    while(1);
    return 0;
}

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define ls(x) (x->l)
#define rs(x) (x->r)
#define sz(x) ((x)?(x->size):(0))
using namespace std;
struct Treap
{
    struct treap
    {
        int size,fix,val;
        treap *l,*r;
        treap(int x)
        {
            val=x;fix=rand();size=1;
            l=r=NULL;
        }
    }*root;
    void pushup(treap *x){x->size=sz(ls(x))+sz(rs(x))+1;}
    void lturn(treap *&x)
    {
        treap *y=rs(x);
        rs(x)=ls(y);pushup(x);
        ls(y)=x;pushup(y);
        x=y;
    }
    void rturn(treap *&x)
    {
        treap *y=ls(x);
        ls(x)=rs(y);pushup(x);
        rs(y)=x;pushup(y);
        x=y;
    }
    void add(treap *&x,int val)
    {
        if(x==NULL)
        {
            x=new treap(val);
            return ;
        }
        if(val<x->val)
        {
            add(ls(x),val);
            pushup(x);
            if(ls(x)->fix>x->fix)rturn(x);
        }
        else
        {
            add(rs(x),val);
            pushup(x);
            if(rs(x)->fix>x->fix)lturn(x);
        }
    }
    void del(treap *&x,int val)
    {
        if(val==x->val)
        {
            if(ls(x)&&rs(x))
            {
                if(ls(x)->fix>rs(x)->fix){rturn(x);del(rs(x),val);}
                else {lturn(x);del(ls(x),val);}
            }
            else
            {
                treap *y=NULL;
                if(ls(x))y=ls(x);
                else y=rs(x);
                delete x;x=y;
            }
        }
        else
        {
            if(val<x->val)del(ls(x),val);
            else del(rs(x),val);
        }
        if(x)pushup(x);
    }
    int rank(int val)
    {
        treap *x=root;
        int ans=0;
        while(x)
        {
            if(val>x->val){ans+=(sz(ls(x))+1);x=rs(x);}
            else x=ls(x);
        }
        return ans;
    }
    treap* kth(int k)
    {
        treap *x=root;
        while(x)
        {
            int size=sz(ls(x))+1;
            if(size==k)return x;
            if(size>=k)x=ls(x);
            // = 向右走，因为add的时候就是向右 
            else {k-=size;x=rs(x);}
        }
    }
}T;

int m,u,o;

int main(){
    //freopen("phs.in","r",stdin);
    //freopen("phs.out","w",stdout);
    scanf("%d",&m);
    while(m--)
    {
        scanf("%d%d",&u,&o);
        if(u==1)T.add(T.root,o);
        else if(u==2)T.del(T.root,o);
        else if(u==3)printf("%d
",T.rank(o)+1);
        else if(u==4)printf("%d
",T.kth(o)->val);
        else if(u==5)printf("%d
",T.kth(T.rank(o))->val);
        else printf("%d
",T.kth(T.rank(o+1)+1)->val);
        //加2有可能返回一个空指针，炸掉你 
    }
    while(1);
    return 0;
}