首先我寫了個凸包就溜了

這是最小圓覆蓋問題，今晚學了一下

先隨機化點，一個個加入

假設當前圓心為o，半徑為r，加入的點為i

若i不在圓裏面，令圓心為i，半徑為0

再重新從1~i-1不停找j不在圓裏面，令圓心為ij中點，直徑為ij距離

再重新在1~j-1不停找k不在圓裏面，三點可確定一圓，初中數學

復雜度看似O(n^3)實則O(n)，好玄學

坑點：註意如果用點斜式表示方程有斜率為不存在的情況，需要特判

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include
 
<algorithm>
#include<cmath>
using namespace std;
const double eps=1e-8;
double sqr(double x){return x*x;}

struct point{ double x,y;point(){} point(double X,double Y){x=X,y=Y;} };
double getdis(point p1,point p2){return sqrt(sqr(p1.x-p2.x)+sqr(p1.y-p2.y));}
point  middle(point p1,point p2){
 
return point((p1.x+p2.x)/2,(p1.y+p2.y)/2);}
double slope (point p1,point p2)
{
    if(p2.x==p1.x)return 0;
    return (p2.y-p1.y)/(p2.x-p1.x);
}
double multi(point p1,point p2,point p0)
{
    double x1,y1,x2,y2;
    x1=p1.x-p0.x;
    y1=p1.y-p0.y;
    x2=p2.x-p0.x;
    y2=p2.y-p0.y;
    return x1*y2-x2*y1;
}

 
struct segment{ double k,b;segment(){} segment(double K,double B){k=K,b=B;} };
segment getseg(double k,point pp){return segment(k,pp.y-k*pp.x);}
point intersection(segment s1,segment s2)
{
    double x=(s2.b-s1.b)/(s1.k-s2.k);
    double y=s1.k*x+s1.b;
    return point(x,y);
}

//--------------------------------------simple--------------------------------------------------------

int n; point p[1100000];
bool cmp(point p1,point p2)
{
    double d=multi(p1,p2,p[1]);
    if(fabs(d)<=eps)return getdis(p1,p[1])<getdis(p2,p[1]);
    else return d>0;
}
int top,sta[1100000];
void graham()
{
    sort(p+2,p+n+1,cmp);
    top=0;sta[++top]=1,sta[++top]=2;
    double g;
    for(int i=3;i<=n;i++)
    {
        while(top>=2)
        {
            g=multi(p[sta[top]],p[i],p[sta[top-1]]);
            if(g<0||fabs(g)<=eps)top--;
            else break;
        }
        sta[++top]=i;
    }
}

//------------------------------------graham----------------------------------------------------------

point getcore(point p1,point p2,point p3)
{
    double g=multi(p1,p2,p3);
    if(fabs(g)<=eps)
    {
        double d1=getdis(p1,p2),d2=getdis(p1,p3),d3=getdis(p2,p3);
        if(d1>d2&&d1>d3)return middle(p1,p2);
        if(d2>d1&&d2>d3)return middle(p1,p3);
        if(d3>d1&&d3>d2)return middle(p2,p3);
    }
    else
    {
        segment s1,s2;
        if(slope(p1,p2)==0)
        {
            s1=getseg(-1/slope(p1,p3),middle(p1,p3));
            s2=getseg(-1/slope(p2,p3),middle(p2,p3));
        }
        else if(slope(p1,p3)==0)
        {
            s1=getseg(-1/slope(p1,p2),middle(p1,p2));
            s2=getseg(-1/slope(p2,p3),middle(p2,p3));
        }
        else
        {
            s1=getseg(-1/slope(p1,p2),middle(p1,p2));
            s2=getseg(-1/slope(p1,p3),middle(p1,p3));
        }
        return intersection(s1,s2);
    }
}

void circlecover()
{
    random_shuffle(sta+1,sta+top+1);
    point o=p[sta[1]];double r=0,d;
    for(int i=2;i<=top;i++)
        if(getdis(o,p[sta[i]])>r)
        {
            o=p[sta[i]],r=0;
            for(int j=1;j<i;j++)
                if(getdis(o,p[sta[j]])>r)
                {
                    o=middle(p[sta[i]],p[sta[j]]),r=getdis(o,p[sta[i]]);
                    for(int k=1;k<j;k++)
                        if(getdis(o,p[sta[k]])>r)
                            o=getcore(p[sta[i]],p[sta[j]],p[sta[k]]),r=getdis(o,p[sta[i]]);
                }
        }
    printf("%.2lf %.2lf %.2lf\n",o.x,o.y,r);
}

//------------------------------------solve----------------------------------------------------------

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf%lf",&p[i].x,&p[i].y);
        if(p[i].y<p[1].y||(p[i].y==p[1].y&&p[i].x<p[1].x))
            swap(p[i],p[1]);
    }
    graham();
    circlecover();
    
    return 0;
}

bzoj2823: [AHOI2012]信號塔&&1336: [Balkan2002]Alien最小圓覆蓋&&1337: 最小圓覆蓋