好难写啊！
这题如果保证数据随机，那么可以直接跑一个最小圆覆盖，先二分半径，再贪心覆盖。
但是出题人显然不会这么善良。
于是就可以倍增，\(1,2,4,8,16...\)，这样尝试长度，找到最大可行二进制长度（即最高位）后，再逐位确定。
复杂度\(O(nlog^2(n))\)
但是写完之后又被卡了精度，改随机数种子才可以过。
#include <bits/stdc++.h>
using namespace std;
const int N=100010;
typedef double ld;
const ld EPS=1e-8;
struct point{
ld x,y;
ld distance(){
return sqrt(x*x+y*y);
}
point operator *(const ld &z) const{
return {z*x,z*y};
}
point operator /(const ld &z) const{
return {x/z,y/z};
}
ld operator ^(const point &_) const{
return x*_.x+y*_.y;
}
ld operator *(const point &_) const{
return x*_.y-y*_.x;
}
point operator +(const point &_) const{
return {x+_.x,y+_.y};
}
ld sqr() const{
return x*x+y*y;
}
point operator -(const point &_) const{
return {x-_.x,y-_.y};
}
point rotate(const ld &alpha) const{
ld tc=cos(alpha),ts=sin(alpha);
return {x*tc-y*ts,x*ts+y*tc};
}
point normal() const{
return {-y,x};
}
}a[N],c[N];
struct line{
point x,y;
point generate(const ld &c) const{
return x+y*c;
}
point cross(const line &u) const{
ld s1=y*u.y;
ld s2=u.y*(x-u.x);
//cerr<<s1<<" "<<s2<<" "<<u.y.x<<" "<<u.y.y<<endl;
//cerr<<x.x<<" "<<x.y<<" "<<y.x<<" "<<y.y<<endl;
return generate(s2/s1);
}
};
struct cir{
point o;
ld r;
bool in(const point &d) const{
return (o-d).sqr()<=r;
}
};
bool rec,re;
int n,m;
int num;
point out[N];
cir solve(const point &x,const point &y,const point &z){
//line l1{(x+y)/2,(x-y).rotate(M_PI_2)};
//line l2{(y+z)/2,(z-y).rotate(M_PI_2)};
line l1{(x+y)/2,(x-y).normal()};
line l2{(y+z)/2,(z-y).normal()};
if (abs(l1.y*l2.y)<EPS){
ld len=max(max((x-y).sqr(),(x-z).sqr()),(y-z).sqr());
if (len==(x-y).sqr()){
return {(x+y)/2,sqrt(len)/2};
}
if (len==(x-z).sqr()){
return {(x+z)/2,sqrt(len)/2};
}
return {(y+z)/2,len/2};
}
point tmp=l1.cross(l2);
return {tmp,(tmp-x).sqr()};
}
ld check(int l,int r){
if (r-l+1==1){
if (re) out[++num]=a[l];
return 0;
}
for (int i=l; i<=r; ++i) c[i]=a[i];
random_shuffle(c+l,c+r+1);
cir tmp={c[l],0};
for (int i=l+1; i<=r; ++i){
if (tmp.in(c[i])) continue;
tmp={c[i],0};
for (int j=l; j<i; ++j)
if (!tmp.in(c[j])){
tmp={(c[i]+c[j])/2,(c[i]-c[j]).sqr()/4};
for (int k=l; k<j; ++k)
if (!tmp.in(c[k])){
//cerr<<i<<" "<<j<<" "<<k<<endl;
//cerr<<c[i].x<<" "<<c[i].y<<endl;
//cerr<<c[j].x<<" "<<c[j].y<<endl;
//cerr<<c[k].x<<" "<<c[k].y<<" "<<l<<" "<<r<<endl;
tmp=solve(c[i],c[j],c[k]);
}
}
}
if (re){
out[++num]=tmp.o;
}
return tmp.r;
}
int getnext(const int &x,const ld &mid){
int i;
for (i=2; x+i-1<=n; i<<=1) if (check(x,x+i-1)>mid*mid) break;
for (int j=(i>>=1); j; j>>=1)
if (x+i+j-1<=n&&check(x,x+i+j-1)<=mid*mid) i+=j;
if (rec){
re=1;
check(x,x+i-1);
re=0;
}
return x+i;
}
bool maybe(const ld &mid){
//cerr<<"maybe:"<<mid<<endl;
int rest=m;
for (int i=1; i<=n; i=getnext(i,mid))
if (--rest<0) return 0;
//cerr<<"SUCC"<<endl;
return 1;
}
int main(){
srand(19260817);
scanf("%d%d",&n,&m);
for (int i=1; i<=n; ++i) scanf("%lf%lf",&a[i].x,&a[i].y);
ld ret=-1;
for (ld l=0,r=3e6,mid=(l+r)/2; l+EPS<r; mid=(l+r)/2)
if (maybe(mid)) r=ret=mid; else l=mid;
cout<<fixed<<setprecision(10);
cout<<ret<<endl;
rec=1;
maybe(ret);
cout<<num<<endl;
for (int i=1; i<=num; ++i) cout<<out[i].x<<" "<<out[i].y<<'\n';
}