black box

topic

topic

solution

Isn't this just using a balanced tree?

How could such a complicated code be used! ! ! (╯‵□′)╯︵┻━┻

It’s really not possible to use discretization and chairman tree.

How is it possible to hit the line segment tree (╯‵□′)╯︵┻━┻

Adding two points to the tree array is fine, you don’t dislike the direct tree array.

I don’t want to fight either (╯‵□′)╯︵┻━┻

Does the balance tree have STL?

No STL(╯‵□′)╯︵┻━┻

Then use the overall dichotomy

No (╯‵□′)╯︵┻━┻

Then learn later.

Here, let’s think about it carefully, this is actually asking $k\; is$ big, but this$k$ turns out to be incremental, which means we can use the heap, and then query the firstkkevery time$k\; is$ big, then kick out the top of the pile, so$k+1$ is the top of the pile again! ! ! ! perfect.

Then I was WA. . . .

Think about it carefully. Although the element on the top of the heap is unlikely to appear again in continuous inquiries, if you finish the inquiries $After\; k\; is$ big, insert a smaller than it to replace the dead ghost, so next time we ask$k+1$ is still the top of the pile.

So we need a tool stack that can be regretted. Yes, it is the top stack.

We put all the deleted numbers in a big root pile, if the number xx inserted at this time$x\; is$ smaller than Duitou, then put$x\; is$ inserted into the big root pile, and the original top of the big root pile is inserted back into the small root pile to participate in the inquiry.

NB on the top pile! ! ! !

Of course, I do it by hand, the code is so long (╯‵□′)╯︵┻━┻.

Code

$O\left(nlogn\right)$

#include<cstdio>
#include<cstring>
#define  N  31000
using  namespace  std;
int  a[N],n,m,b[N];
inline  void  wap(int  &x,int  &y){
    
    int  t=x;x=y;y=t;}
struct  xiao_dui
{
    
    
	int  a[N],tot;
	void  ins(int  x)
	{
    
    
		a[++tot]=x;int  now=tot;
		while(now>1)
		{
    
    
			int  i=now>>1;
			if(a[i]>a[now])wap(a[i],a[now]),now=i;
			else  break;
		}
	}
	int  pop()
	{
    
    
		int  ans=a[1];wap(a[1],a[tot]);tot--;
		int  now=1;
		while((now<<1)<=tot)
		{
    
    
			int  i=now<<1;
			if(i<tot  &&  a[i+1]<a[i])i++;
			if(a[i]<a[now])wap(a[i],a[now]),now=i;
			else  break;
		}
		return  ans;
	}
}a1;
struct  da_dui
{
    
    
	int  a[N],tot;
	void  ins(int  x)
	{
    
    
		a[++tot]=x;int  now=tot;
		while(now>1)
		{
    
    
			int  i=now>>1;
			if(a[i]<a[now])wap(a[i],a[now]),now=i;
			else  break;
		}
	}
	int  pop()
	{
    
    
		int  ans=a[1];wap(a[1],a[tot]);tot--;
		int  now=1;
		while((now<<1)<=tot)
		{
    
    
			int  i=now<<1;
			if(i<tot  &&  a[i+1]>a[i])i++;
			if(a[i]>a[now])wap(a[i],a[now]),now=i;
			else  break;
		}
		return  ans;
	}
	int  top(){
    
    return  tot==0?-999999999:a[1];}
}a2;
int  main()
{
    
    
	scanf("%d%d",&n,&m);
	for(int  i=1;i<=n;i++)scanf("%d",&a[i]);
	for(int  i=1;i<=m;i++)
	{
    
    
		scanf("%d",&b[i]);
		b[i]=b[i]-i+1;
	}
	int  t=0;
	for(int  i=1;i<=n;i++)
	{
    
    
		if(a[i]<a2.top())
		{
    
    
			a1.ins(a2.pop());
			a2.ins(a[i]);
		}
		else  a1.ins(a[i]);
		while(a1.tot==b[t+1]  &&  t<m)
		{
    
    
			t++;
			int  x=a1.pop();a2.ins(x);
			printf("%d\n",x);
		}
		if(t==m)break;
	}
	return  0;
}