有一个点超时，确实是个很简单的splay #include
    
     #include
     
      using namespace std;int n,shu[1000006][2],root,size,b1,b2,sum1,sum[1000005],zhi[1000005];int fa[1000005];void xuan(int a1){	int a2,a3,l,r;	a2=fa[a1];	a3=fa[a2];	if(a2==root)	  root=a1;	else	  if(a2==shu[a3][0])	    shu[a3][0]=a1;	  else	    shu[a3][1]=a1;	if(shu[a2][0]==a1)	  l=0;	else	  l=1;	r=l^1;	fa[a1]=a3;	fa[a2]=a1;	shu[a2][l]=shu[a1][r];	shu[a1][r]=a2;	fa[shu[a2][l]]=a2;	return;}void zhuan(int a1){	int a2,a3;	for(;a1!=root;)	  {	  	a2=fa[a1];	  	a3=fa[a2];	  	if(a2!=root)	  	  if(a2==shu[a3][0]^a1==shu[a2][0])	  	    xuan(a1);	  	  else	  	    xuan(a2);	  	xuan(a1);	  }}void jia(int &a1,int a2,int a3){	if(a1==0)	  {	  	size++;	  	a1=size;	  	zhi[a1]=a2;	  	sum[a1]=1;	  	fa[a1]=a3;	  	zhuan(a1);	  	return;	  }	if(zhi[a1]==a2)	  sum[a1]++;	else	  if(zhi[a1]
      
       a2)	    {	    	b2=zhi[a1];	    	hou(shu[a1][0],a2);		}	  else	    hou(shu[a1][1],a2);	return;}int main(){	scanf("%d",&n);	scanf("%d",&sum1);	jia(root,sum1,0);	for(int i=1;i