A - Archery Tournament dynamic prescription + vecotor immortal title

Like rectangles of deposit or deposit diagram, round numbers stored within each node where the district level, then violence judge, I began to have this idea, however. . . This TM could go. . . Think about it, looks like you can over time complexity of metaphysics can not prove. . . .

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<set>
#include<vector>
#define LL long long
using namespace std;
const int maxx = 2e5+6;
vector<int>v[maxx*20];
int xx[maxx];
int yy[maxx];
struct node
{
    int l,r;
} tree[maxx*20];
int cnt;
int ans;
bool check(int x1,int y1,int x2,int y2)
{
    return 1LL*(x1-x2)*(x1-x2)+1LL*(y1-y2)*(y1-y2)<1LL*y1*y1;
}
void buildtree(int &rt,int l,int r,int L,int R,int id)
{
    if (!rt)
    {
        cnt++;
        rt=cnt;
    }
    if (L<=l && r<=R)
    {
        v[rt].push_back(id);
        return ;
    }
    int mid=(l+r)>>1;
    if (L<=mid)
    {
        buildtree(tree[rt].l,l,mid,L,R,id);
    }
    if (R>mid)
    {
        buildtree(tree[rt].r,mid+1,r,L,R,id);
    }
}
void query(int rt,int l,int r,int x,int y)
{
    if (!rt)return;
    for (auto it:v[rt])
    {
        if(check(xx[it],yy[it],x,y))
        {
            ans=it;
            return ;
        }
    }
    if (l==r)
    {
        return ;
    }
    int mid=(l+r)>>1;
    if (x<=mid)
    {
        query(tree[rt].l,l,mid,x,y);
    }
    else
    {
        query(tree[rt].r,mid+1,r,x,y);
    }
}
void update(int rt,int l,int r,int ul,int ur,int id)
{
    if (ul>ur)return;
    if (ul<=l && r<=ur)
    {
        vector<int>tmp;
        for (auto it:v[rt])
        {
            if (it!=id)
            {
                tmp.push_back(it);
            }
        }
        v[rt]=tmp;
        return ;
    }
    int mid=(l+r)>>1;
    if (ul<=mid)
    {
        update(tree[rt].l,l,mid,ul,ur,id);
    }
    if (ur>mid)
    {
        update(tree[rt].r,mid+1,r,ul,ur,id);
    }
}
int main()
{
    int t,op;
    int rt;
    while(~scanf("%d",&t))
    {
        cnt=0;
        rt=0;
        for (int i=1; i<=t; i++)
        {
            scanf("%d%d%d",&op,&xx[i],&yy[i]);
            if (op==1)
            {
                buildtree(rt,-1e9,1e9,xx[i]-yy[i],xx[i]+yy[i],i);
            }
            else
            {
                ans=-1;
                query(rt,-1e9,1e9,xx[i],yy[i]);
                printf("%d\n",ans);
                if (ans!=-1)
                {
                    update(rt,-1e9,1e9 xx [years] -yy [years] xx [years] + yy [years] years); 
                } 
            } 
        } 
    } 
    Return  0 ; 
}