BZOJ1336&1337：[Balkan2002]Alien最小圆覆盖

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题目大意：给定n个点，求最小圆覆盖

最小圆覆盖裸题....算法讲解戳→这里

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define N 100010using namespace std;struct node{double x,y;}b[N];node O;double R;double sqr(double x){return x*x;}double dis(node x,node y){    return sqrt(sqr(x.x-y.x)+sqr(x.y-y.y));}bool incircle(node x){    if(dis(O,x)<=R) return true;    return false;}node solve(double a,double b,double c,double d,double e,double f){    double y=(f*a-c*d)/(b*d-e*a);    double x=(f*b-c*e)/(a*e-b*d);    return (node){x,y};}int main(){    int n;    scanf("%d",&n);    int i,j,k;    for(i=1;i<=n;i++)    scanf("%lf%lf",&b[i].x,&b[i].y);    random_shuffle(b+1,b+n+1);    R=0;    for(i=1;i<=n;i++)    if(!incircle(b[i]))    {        O.x=b[i].x;O.y=b[i].y;R=0;        for(j=1;j<i;j++)        if(!incircle(b[j]))        {            O.x=(b[i].x+b[j].x)/2;            O.y=(b[i].y+b[j].y)/2;            R=dis(O,b[i]);            for(k=1;k<j;k++)            if(!incircle(b[k]))            {                O=solve(                b[i].x-b[j].x,b[i].y-b[j].y,(sqr(b[j].x)+sqr(b[j].y)-sqr(b[i].x)-sqr(b[i].y))/2,                b[i].x-b[k].x,b[i].y-b[k].y,(sqr(b[k].x)+sqr(b[k].y)-sqr(b[i].x)-sqr(b[i].y))/2                 );                R=dis(b[i],O);            }        }    }    printf("%.10lf\n%.10lf %.10lf",R,O.x,O.y);}

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