[TYVJ 1730]二逼平衡树 线段树套平衡树

来一发树套树。1A也是很感动QAQ

就是时间复杂度略大。而且好像还有其他打法。

谨以此纪念此类型树套树入门

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
#define pos(i,a,b) for(int i=(a);i<=(b);i++)
#define N 51000
#define size(x) ((x)?(x->size):(0))
#define lc(x) (x->ch[0])
#define rc(x) (x->ch[1])
int n,m;
struct Treap{
	Treap *ch[2];
	int v,key,size;
	Treap(int x=0){
		ch[0]=ch[1]=NULL;key=rand();size=1;v=x;
	}
}*root[N*4];
void pushup(Treap *rt){
	rt->size=size(lc(rt))+size(rc(rt))+1;
}
void turn(Treap *&rt,int d){
	Treap *t=rt->ch[d^1];
	rt->ch[d^1]=t->ch[d];
	pushup(rt);
	t->ch[d]=rt;
	pushup(t);
	rt=t;
}
void insert(Treap *&rt,int x){
	if(!rt){
		rt=new Treap(x);return;
	}
	int d=x < rt->v;
	insert(rt->ch[d^1],x);
	pushup(rt);
	if(rt->key > rt->ch[d^1]->key) turn(rt,d);
}
void del(Treap *&rt,int x){
	if(rt->v == x){
		if(lc(rt)&&rc(rt)){
			int d=lc(rt)->key < rc(rt)->key;
			turn(rt,d);
			del(rt->ch[d],x);
		}
		else{
			Treap *t=NULL;
			if(lc(rt)) t=lc(rt);
			else t=rc(rt);
			delete rt;rt=t;
		}
	}
	else{
		int d=rt->v < x;
		del(rt->ch[d],x);
	}
	if(rt) pushup(rt);
}
int pai(Treap *&rt,int x){
	if(!rt) return 0;
	if(x<= rt->v){
		return pai(lc(rt),x);
	}
	else return size(lc(rt))+1+pai(rc(rt),x);
}
int a[N];
void Insert(int l,int r,int rt){
	pos(i,l,r) insert(root[rt],a[i]);
}
void build(int l,int r,int rt){
	Insert(l,r,rt);
	if(l==r) return;
	int mid=(l+r)>>1;
	build(l,mid,rt<<1);
	build(mid+1,r,rt<<1|1);
}
void update(int l,int r,int pos,int x,int y,int rt){
	del(root[rt],x);insert(root[rt],y);
	if(l==r) return;
	int mid=(l+r)>>1;
	if(pos<=mid) update(l,mid,pos,x,y,rt<<1);
	else update(mid+1,r,pos,x,y,rt<<1|1);
}
int Pai(int xl,int xr,int k,int l,int r,int rt){
	if(l>=xl&&r<=xr) return pai(root[rt],k);
	int mid=(l+r)>>1;
	int ans(0);
	if(xl<=mid) ans+=Pai(xl,xr,k,l,mid,rt<<1);
	if(xr>mid) ans+=Pai(xl,xr,k,mid+1,r,rt<<1|1);
	return ans;
}
int kth(int L,int R,int k){
	int l=0,r=(int)1e8;
	while(l<=r){
		int mid=(l+r)>>1;
		if(Pai(L,R,mid,1,n,1)+1<=k) l=mid+1;
		else r=mid-1;
	}
	return r;
}
int main(){
	scanf("%d%d",&n,&m);
	pos(i,1,n) scanf("%d",&a[i]);
	build(1,n,1);
	pos(i,1,m){
		int opt;scanf("%d",&opt);
		int l,r,k;
		if(opt==3){
			scanf("%d%d",&l,&k);
			update(1,n,l,a[l],k,1);
			a[l]=k;
			continue;
		}
		scanf("%d%d%d",&l,&r,&k);
		if(opt==1) printf("%d\n",Pai(l,r,k,1,n,1)+1);
		else if(opt==2) printf("%d\n",kth(l,r,k));
		else if(opt==4) printf("%d\n",kth(l,r,Pai(l,r,k,1,n,1)));
		else if(opt==5) printf("%d\n",kth(l,r,Pai(l,r,k+1,1,n,1)+1));
	}
	return 0;
}