LOJ#6360. 复燃「恋之埋火」（最小圆覆盖+高斯消元）

题面

传送门

题解

不难发现最小圆覆盖的随机增量法复杂度还是正确的

所以现在唯一的问题就是给定若干个点如何求一个\(m\)维的圆

其实就是这一题

//minamoto
#include<bits/stdc++.h>
#define R register
#define inline __inline__ __attribute__((always_inline))
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R int res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
double readdb()
{
    R double x=0,y=0.1,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(x=ch-'0';(ch=getc())>='0'&&ch<='9';x=x*10+ch-'0');
    for(ch=='.'&&(ch=getc());ch>='0'&&ch<='9';x+=(ch-'0')*y,y*=0.1,ch=getc());
    return x*f;
}
const int N=2e4+5;const double eps=1e-8;
inline int sgn(double x){return x<-eps?-1:x>eps;}
int n,k;
struct node{
	double a[9];
	inline double &operator [](const int &x){return a[x];}
	inline node operator +(node &b)const{
		node c;
		fp(i,1,k)c[i]=a[i]+b[i];
		return c;
	}
	inline node operator -(node &b)const{
		node c;
		fp(i,1,k)c[i]=a[i]-b[i];
		return c;
	}
	inline double operator ^(node &b)const{
		double res=0;
		fp(i,1,k)res+=a[i]*b[i];
		return res;
	}
	inline node operator *(const double &b)const{
		node c;
		fp(i,1,k)c[i]=a[i]*b;
		return c;
	}
	inline double norm()const{
		double res=0;
		fp(i,1,k)res+=a[i]*a[i];
		return res;
	}
}p[N],c[N],o;
double a[9][9],r;
void Gauss(int n){
	fp(i,1,n){
		int k=i;
		fp(j,i+1,n)if(fabs(a[j][i])>fabs(a[k][i]))k=j;
		if(i!=k)fp(j,i,n+1)swap(a[i][j],a[k][j]);
		double t=1.0/a[i][i];
		fp(j,i,n+1)a[i][j]*=t;
		fp(j,i+1,n)fd(k,n+1,i)a[j][k]-=a[j][i]*a[i][k];
	}
	fd(i,n,1){
		if(!sgn(a[i][i])){a[i][i]=0;continue;}
		double t=1.0/a[i][i];
		fd(j,i-1,1)fd(k,n+1,i)a[j][k]-=a[i][k]*t*a[j][i];
		a[i][n+1]*=t;
	}
}
void circle(int n){
	if(n==0)return o=c[1],r=0,void();
	if(n==1)return o=c[1],r=0,void();
	if(n==2)return o=(c[1]+c[2])*0.5,r=(o-c[1]).norm(),void();
	fp(i,2,n)c[i]=c[i]-c[1];
	fp(i,2,n)fp(j,i,n)a[i-1][j-1]=a[j-1][i-1]=(c[i]^c[j])*2;
	fp(i,2,n)a[i-1][n]=c[i]^c[i];
	Gauss(n-1);
	o=c[1];
	fp(i,2,n)o=c[i]*a[i-1][n]+o;
	r=(o-c[1]).norm();
	fp(i,2,n)c[i]=c[i]+c[1];
}
void solve(int n,int cnt){
	circle(cnt);
	fp(i,1,n)if(sgn((p[i]-o).norm()-r)>0)
		c[cnt+1]=p[i],solve(i-1,cnt+1);
}
int main(){
//	freopen("testdata.in","r",stdin);
	srand(20030719);
	n=read(),k=read();
	fp(i,1,n)fp(j,1,k)p[i][j]=readdb();
	random_shuffle(p+1,p+1+n);
	solve(n,0);
	fp(i,1,k)printf("%.6lf%c",o[i]," \n"[i==k]);
	return 0;
}