有旋treap求调,原版能过QwQ
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有旋treap求调,原版能过QwQ
523541
wdy1028楼主2021/10/7 15:03

样例输出5,交上去有20分,吐了

#include <bits/stdc++.h>
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
using namespace std;
const int INF = 2147483647;
const int S = 1e6+5;
struct treap
{
	int l,r;
	int val,dat;
	int cnt,size;
}a[S];
int tot,root,last,n,m,tmp,ans;
inline int New(int val)
{
	a[++tot].val = val;
	a[tot].dat = rand();
	a[tot].cnt = a[tot].size = 1;
	return tot;	
}
inline void Update(int p){a[p].size = a[a[p].l].size + a[a[p].r].size + a[p].cnt;}
inline int Get1(int p,int val)
{
	if(!p) return 0;
	if(val == a[p].val) return a[a[p].l].size + 1;
	if(val < a[p].val) return Get1(a[p].l,val);
	return Get1(a[p].r,val) + a[a[p].l].size + a[p].cnt; 	
}
inline int Get2(int p,int rank)
{
	if(!p) return INF;
	if(a[a[p].l].size >= rank) return Get2(a[p].l,rank);
	if(a[a[p].l].size + a[p].cnt >= rank) return a[p].val;
	return Get2(a[p].r,rank - a[a[p].l].size - a[p].cnt);	
}
inline void zig(int &p)
{
	int q = a[p].l;
	a[p].l = a[q].r,a[q].r = p,p = q;
	Update(a[p].r),Update(p);
}
inline void zag(int &p)
{
	int q = a[p].r;
	a[p].r = a[q].l,a[q].l = p,p = q;
	Update(a[p].l),Update(p);
}
inline void Insert(int &p,int val)
{
	if(!p)
	{
		p = New(val);
		return;		
	}	
	if(val == a[p].val)
	{
		a[p].cnt++,Update(p);
		return;
	}
	if(val < a[p].val)
	{
		Insert(a[p].l,val);
		if(a[p].dat < a[a[p].l].dat) zig(p);
	}
	else
	{
		Insert(a[p].r,val);
		if(a[p].dat < a[a[p].r].dat) zag(p);		
	}
	Update(p);	
}
inline int Getpre(int val)
{
	int ans = 1,p = root;
	while(p)
	{
		if(val == a[p].val)
		{
			if(a[p].l > 0)
			{
				p = a[p].l;
				while(a[p].r > 0) p = a[p].r;
				ans = p;
			}
			break;
		}
		if(a[p].val < val && a[p].val > a[ans].val) ans = p;
		p = val<a[p].val?a[p].l:a[p].r;
	}
	return a[ans].val;
}
inline int Getnext(int val)
{
	int ans = 2,p = root;
	while(p)
	{
		if(val == a[p].val)
		{
			if(a[p].r > 0)
			{
				p = a[p].r;
				while(a[p].l > 0) p = a[p].l;
				ans = p;
			}
			break;
		}
		if(a[p].val > val && a[p].val < a[ans].val) ans = p;
		p = val<a[p].val?a[p].l:a[p].r;
	}
	return a[ans].val;	
}
inline void Remove(int &p,int val)
{
	if(!p) return;
	if(val == a[p].val)
	{
		if(a[p].cnt > 1)
		{
			--a[p].cnt;
			Update(p);
			return;
		}
		if(a[p].l || a[p].r)
		{
			if(!a[p].r || a[a[p].l].dat > a[a[p].r].dat) zig(p),Remove(a[p].r,val);
			else zag(p),Remove(a[p].l,val);
			Update(p);
		}
		else p = 0;
		return;
	}
	val < a[p].val?Remove(a[p].l,val):Remove(a[p].r,val);
	Update(p);	
}
int main(int argc,char *argv[])
{
	New(-INF),New(INF);
	root = 1;a[1].r = 2;
	scanf("%d%d",&n,&m);
	for(int i = 1;i <= n;++i) 
	{
		scanf("%d",&tmp);
		Insert(root,tmp);
	}
	while(m--)
	{
		int opt,x;
		scanf("%d%d",&opt,&x);
		x ^= last;
		switch(opt)
		{
			case 1:
				Insert(root,x);break;
			case 2:
				Remove(root,x);break;
			case 3:
				last = Get1(root,x)-1;
				ans ^= last;
				break;
			case 4:
				last = Get2(root,x+1);
				ans ^= last;
				break;
			case 5:
				last = Getpre(x);
				ans ^= last;
				break;
			case 6:
				last = Getnext(x);
				ans ^= last;
				break;
		}
	}
	printf("%d\n",ans);
	return 0;
}
	
2021/10/7 15:03
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