【线段树套平衡树】【pb_ds】bzoj3196 Tyvj 1730 二逼平衡树

线段树套pb_ds里的平衡树,在洛谷OJ上测试,后三个测试点TLE

#include<cstdio>
#include<algorithm>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define N 50010
#define INF 2147483647
using namespace std;
using namespace __gnu_pbds;
typedef pair<int,int> Point;
typedef tree<Point,null_type,less<Point>,rb_tree_tag,tree_order_statistics_node_update> rb_tree;
typedef rb_tree::iterator ITER;
rb_tree T[N<<2];
int a[N],tags[N],tag,n,m;
void buildtree(int rt,int l,int r)
{
	if(l==r)
	  {
	  	T[rt].insert(Point(a[l],l));
	  	return;
	  }
	int m=(l+r>>1);
	buildtree(rt<<1,l,m);
	buildtree(rt<<1|1,m+1,r);
	T[rt]=T[rt<<1];
	for(ITER it=T[rt<<1|1].begin();it!=T[rt<<1|1].end();++it)
	  T[rt].insert(*it);
}
int Rank(int v,int TAG,int ql,int qr,int rt,int l,int r)
{
	if(ql<=l&&r<=qr)
	  return T[rt].order_of_key(Point(v,TAG));
	int m=(l+r>>1),res=0;
	if(ql<=m) res+=Rank(v,TAG,ql,qr,rt<<1,l,m);
	if(m<qr) res+=Rank(v,TAG,ql,qr,rt<<1|1,m+1,r);
	return res;
}
int Pre(int v,int ql,int qr,int rt,int l,int r)
{
	if(ql<=l&&r<=qr)
	  {
	  	ITER it=T[rt].lower_bound(Point(v,0));
	  	if(it==T[rt].begin()) return -INF;
		--it;
	  	return (*it).first;
	  }
	int m=(l+r>>1),res=-INF;
	if(ql<=m) res=max(res,Pre(v,ql,qr,rt<<1,l,m));
	if(m<qr) res=max(res,Pre(v,ql,qr,rt<<1|1,m+1,r));
	return res;
}
int Nex(int v,int ql,int qr,int rt,int l,int r)
{
	if(ql<=l&&r<=qr)
	  {
	  	ITER it=T[rt].upper_bound(Point(v,INF));
	  	if(it==T[rt].end())
		  return INF;
	  	return (*it).first;
	  }
	int m=(l+r>>1),res=INF;
	if(ql<=m) res=min(res,Nex(v,ql,qr,rt<<1,l,m));
	if(m<qr) res=min(res,Nex(v,ql,qr,rt<<1|1,m+1,r));
	return res;
}
void Update(int p,int v,int rt,int l,int r)
{
	T[rt].erase(Point(a[p],tags[p]));
	T[rt].insert(Point(v,tag));
	if(l==r) return;
	int m=(l+r>>1);
	if(p<=m) Update(p,v,rt<<1,l,m);
	else Update(p,v,rt<<1|1,m+1,r);
}
int Kth(int K,int ql,int qr)
{
	int l=0,r=100000000;
	while(l<r)
	  {
	  	int m=(l+r>>1);
	  	if(Rank(m,INF,ql,qr,1,1,n)>=K)
	  	  r=m;
	  	else
	  	  l=m+1;
	  }
	return l;
}
int main()
{
//	freopen("bzoj3196.in","r",stdin);
	int op,x,y,z;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)
	  {
	  	scanf("%d",&a[i]);
	  	tags[i]=i;
	  }
	tag=n;
	buildtree(1,1,n);
//	for(int i=1;i<=n*4;++i)
//	  printf("%d ",(*T[i].begin()).second);
//	puts("");
	for(;m;--m)
	  {
	  	scanf("%d%d%d",&op,&x,&y);
	  	if(op==1)
	  	  {
	  	  	scanf("%d",&z);
	  	  	printf("%d
",Rank(z,0,x,y,1,1,n)+1);
	  	  }
	  	else if(op==2)
	  	  {
	  	  	scanf("%d",&z);
	  	  	printf("%d
",Kth(z,x,y));
	  	  }
	  	else if(op==3)
	  	  {
	  	  	++tag;
	  	  	Update(x,y,1,1,n);
	  	  	a[x]=y;
	  	  	tags[x]=tag;
	  	  }
	  	else if(op==4)
	  	  {
	  	  	scanf("%d",&z);
	  	  	printf("%d
",Pre(z,x,y,1,1,n));
	  	  }
	  	else
	  	  {
	  	  	scanf("%d",&z);
	  	  	printf("%d
",Nex(z,x,y,1,1,n));
	  	  }
	  }
	return 0;
}
原文地址:https://www.cnblogs.com/autsky-jadek/p/5934325.html