BZOJ2626: JZPFAR

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2626

题解：裸K-Dtree，最大值？自己yy一下估价函数就好了。

两题居然是同一个错误，真是too naive。。。

代码：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<string>
 #define inf 1ll<<62
 #define maxn 200000+5
 #define maxm 100000+5
 #define eps 1e-10
 #define ll double
 #define pa pair<ll,int>
 #define for0(i,n) for(int i=0;i<=(n);i++)
 #define for1(i,n) for(int i=1;i<=(n);i++)
 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
 #define mod 1000000007
 #define sqr(x) (x)*(x)
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}
     return x*f;
 }
 int n,m,cur;
 priority_queue<pa,vector<pa>,greater<pa> >q;
 struct rec
 {
     int mi[],mx[],d[],l,r,id;
     int& operator [](int i){return d[i];}
 }p[maxn],t[maxn],now;
 bool operator <(rec a,rec b){return a[cur]<b[cur];}
 inline void pushup(int k)
 {
     int l=t[k].l,r=t[k].r;
     for0(i,)
     {
         t[k].mi[i]=min(t[k][i],min(t[l].mi[i],t[r].mi[i]));
         t[k].mx[i]=max(t[k][i],max(t[l].mx[i],t[r].mx[i]));
     }
 }
 inline int build(int l,int r,int dir)
 {
     int mid=(l+r)>>;
     cur=dir;
     nth_element(p+l,p+mid,p+r+);
     t[mid]=p[mid];
     for0(i,)t[mid].mi[i]=t[mid].mx[i]=t[mid][i];
     t[mid].l=l>mid-?:build(l,mid-,dir^);
     t[mid].r=mid+>r?:build(mid+,r,dir^);
     pushup(mid);
     return mid;
 }
 inline ll calc(int k)
 {
     if(!k)return -inf-;
     ll ret=;
     for0(i,)ret+=max((ll)sqr(now[i]-t[k].mi[i]),(ll)sqr(t[k].mx[i]-now[i]));
     return ret;
 }
 inline ll dist(rec a,rec b){return (ll)sqr(a[]-b[])+(ll)sqr(a[]-b[]);}
 inline void query(int k)
 {
     if(!k)return;
     ll dl=calc(t[k].l),dr=calc(t[k].r),d=dist(t[k],now);
     if(d>q.top().first||(d==q.top().first&&t[k].id<-q.top().second))q.pop(),q.push(pa(d,-t[k].id));
     if(dl>dr)
     {
         if(dl>=q.top().first)query(t[k].l);
         if(dr>=q.top().first)query(t[k].r);
     }else
     {
         if(dr>=q.top().first)query(t[k].r);
         if(dl>=q.top().first)query(t[k].l);
     }
 }

 int main()
 {
     freopen("input.txt","r",stdin);
     freopen("output.txt","w",stdout);
     n=read();
     for1(i,n)p[i][]=read(),p[i][]=read(),p[i].id=i;
     for0(i,)t[].mi[i]=,t[].mx[i]=-;
     int rt=build(,n,);
     m=read();
     while(m--)
     {
         now[]=read();now[]=read();int k=read();
         while(!q.empty())q.pop();
         for1(i,k)q.push(pa(-inf,));
         query(rt);
         printf("%d\n",-q.top().second);
     }
     return ;
 }