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<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,c,root,cnt;
struct point
{
    int d[2];
    bool operator <(const point&a) const
    {
        return d[c]<a.d[c];
    }
}a[N];
struct KDTree{int ch[2],a[2][2];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[0]-v.d[0])+sqr(u.d[1]-v.d[1]);}
ll dis(point u,int a[2][2]){return sqr(max(u.d[0]-a[0][0],a[0][1]-u.d[0]))+sqr(max(u.d[1]-a[1][0],a[1][1]-u.d[1]));}
void build(int &k,int l,int r,int op)
{
    if (l>r) return;
    k=++cnt,c=op;int mid=l+r>>1;nth_element(a+l,a+mid,a+r+1);
    tree[k].p=a[mid];tree[k].a[0][0]=tree[k].a[0][1]=a[mid].d[0],tree[k].a[1][0]=tree[k].a[1][1]=a[mid].d[1];
    for (int i=l;i<=r;i++)
    tree[k].a[0][0]=min(tree[k].a[0][0],a[i].d[0]),tree[k].a[0][1]=max(tree[k].a[0][1],a[i].d[0]),
    tree[k].a[1][0]=min(tree[k].a[1][0],a[i].d[1]),tree[k].a[1][1]=max(tree[k].a[1][1],a[i].d[1]);
    build(tree[k].ch[0],l,mid-1,op^1);
    build(tree[k].ch[1],mid+1,r,op^1);
}
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[0],r=tree[k].ch[1];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()<<1;
    for (int i=1;i<=n;i++) a[i].d[0]=read(),a[i].d[1]=read();
    build(root,1,n,0);
    while (m--) q.push(0);
    for (int i=1;i<=n;i++) query(root,a[i]);
    cout<<q.top();
    return 0;
}