BZOJ4520 CQOI2016K远点对（KD-Tree+堆）

　　堆维护第k大，每个点KD-Tree上A*式查询较远点，跑得飞快，复杂度玄学。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,c,root,cnt;
struct point
{
    int d[];
    bool operator <(const point&a) const
    {
        return d[c]<a.d[c];
    }
}a[N];
struct KDTree{int ch[],a[][];point p;
}tree[N];
priority_queue<ll,vector<ll>,greater<ll> > q;
ll sqr(int x){return 1ll*x*x;}
ll dis(point u,point v){return sqr(u.d[]-v.d[])+sqr(u.d[]-v.d[]);}
ll dis(point u,int a[][]){return sqr(max(u.d[]-a[][],a[][]-u.d[]))+sqr(max(u.d[]-a[][],a[][]-u.d[]));}
void build(int &k,int l,int r,int op)
{
    if (l>r) return;
    k=++cnt,c=op;int mid=l+r>>;nth_element(a+l,a+mid,a+r+);
    tree[k].p=a[mid];tree[k].a[][]=tree[k].a[][]=a[mid].d[],tree[k].a[][]=tree[k].a[][]=a[mid].d[];
    for (int i=l;i<=r;i++)
    tree[k].a[][]=min(tree[k].a[][],a[i].d[]),tree[k].a[][]=max(tree[k].a[][],a[i].d[]),
    tree[k].a[][]=min(tree[k].a[][],a[i].d[]),tree[k].a[][]=max(tree[k].a[][],a[i].d[]);
    build(tree[k].ch[],l,mid-,op^);
    build(tree[k].ch[],mid+,r,op^);
}
void query(int k,point p)
{
    if (dis(tree[k].p,p)>=q.top()) q.push(dis(tree[k].p,p)),q.pop();
    int l=tree[k].ch[],r=tree[k].ch[];ll u=dis(p,tree[l].a),v=dis(p,tree[r].a);
    if (u<v) swap(l,r),swap(u,v);
    if (l&&u>=q.top()) query(l,p);
    if (r&&v>=q.top()) query(r,p);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4520.in","r",stdin);
    freopen("bzoj4520.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read()<<;
    for (int i=;i<=n;i++) a[i].d[]=read(),a[i].d[]=read();
    build(root,,n,);
    while (m--) q.push();
    for (int i=;i<=n;i++) query(root,a[i]);
    cout<<q.top();
    return ;
}