# [luogu3369/bzoj3224]普通平衡树(splay模板、平衡树初探)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<cstdlib>
using namespace std;
typedef long long ll;
const int N = 100005;
int ch[N][2],par[N],val[N],cnt[N],size[N],ncnt,root;

bool chk(int x){
return ch[par[x]][1]==x;
}

void pushup(int x){
size[x]=size[ch[x][0]]+size[ch[x][1]]+cnt[x];//和线段树不同的是需要+上自身的cnt
}

void rotate(int x){
int y=par[x],z=par[y],k=chk(x),w=ch[x][k^1];
ch[y][k]=w;par[w]=y;
ch[z][chk(y)]=x;par[x]=z;
ch[x][k^1]=y;par[y]=x;
pushup(y);pushup(x);
}

void splay(int x,int goal=0){
while(par[x]!=goal){
int y=par[x],z=par[y];
if(z!=goal){
if(chk(x)==chk(y)) rotate(y);
else rotate(x);
}
rotate(x);
}
if(!goal) root=x;
}

void insert(int x){
int cur=root,p=0;
while(cur&&val[cur]!=x){
p=cur;
cur=ch[cur][x>val[cur]];
}
if(cur){
cnt[cur]++;
}else{
cur=++ncnt;
if(p) ch[p][x>val[p]]=cur;
ch[cur][0]=ch[cur][1]=0;
par[cur]=p;val[cur]=x;
cnt[cur]=size[cur]=1;
}
splay(cur);
}

void find(int x){
int cur=root;
if(!cur) return;
while(ch[cur][x>val[cur]]&&x!=val[cur]){
cur=ch[cur][x>val[cur]];
}
splay(cur);
}
//从0开始
int kth(int k){
k++;
int cur=root;
while(1){
if(ch[cur][0]&&k<=size[ch[cur][0]]){
cur=ch[cur][0];
}else if(k>size[ch[cur][0]]+cnt[cur]){
k-=size[ch[cur][0]]+cnt[cur];
cur=ch[cur][1];
}else{
return cur;
}
}
}

int rnk(int x){
find(x);
if(val[root]>=x) return size[ch[root][0]];
else return size[ch[root][0]]+cnt[root];
}

int pre(int x){
find(x);
if(val[root]<x) return root;
int cur=ch[root][0];
while(ch[cur][1]) cur=ch[cur][1];
return cur;
}

int succ(int x) {
find(x);
if(val[root]>x) return root;
int cur=ch[root][1];
while(ch[cur][0]) cur=ch[cur][0];
return cur;
}

void remove(int x) {
int last=pre(x),nxt=succ(x);
splay(last);splay(nxt,last);
int del=ch[nxt][0];
if(cnt[del]>1){
cnt[del]--;
splay(del);
}
else ch[nxt][0]=0;
}

void init(){
insert(-2e9);
insert(2e9);
}

int n,op,x;

int main() {
scanf("%d",&n);
init();
while(n--){
scanf("%d%d",&op,&x);
switch(op){
case 1: insert(x); break;
case 2: remove(x); break;
case 3: printf("%d\n",rnk(x)); break;
case 4: printf("%d\n",val[kth(x)]); break;
case 5: printf("%d\n",val[pre(x)]); break;
case 6: printf("%d\n",val[succ(x)]); break;
}
}
}

posted @ 2019-02-12 03:13  Elpsywk  阅读(145)  评论(0编辑  收藏  举报