BZOJ 2823: [AHOI2012]信号塔

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2823

随机增量法。不断加点维护圆，主要是三点共圆那里打得烦（其实也就是个两中垂线求交点+联立方程求交点而已TAT。。

#include<cstring>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 1000009
#define eps 1e-6
using namespace std;
typedef unsigned long long ll;
struct P{double x,y;
}a[maxn],o;
double r;
int n;
int read(){
    int x=0,f=1; char ch=getchar();
    while (!isdigit(ch)){
        if (ch=='-') f=-1; ch=getchar();
    }
    while (isdigit(ch)){
        x=x*10+ch-'0'; ch=getchar();
    }
    return x*f;
}
P get2(P a,P b){
    P tmp;
    tmp.x=(a.x+b.x)/2; tmp.y=(a.y+b.y)/2;
    return tmp;
}
double sqr(double x){
    return x*x;
}
double dis(P a,P b){
    return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
}
P cross(double a1,double a2,double b1,double b2,double c1,double c2){
    P tmp;
    tmp.x=(c2*b1-c1*b2)/(a1*b2-a2*b1); tmp.y=(c2*a1-c1*a2)/(b1*a2-b2*a1);
    return tmp;
}
P get3(P a,P b,P c){
    P tmp;
    double a1=b.x-a.x,b1=b.y-a.y,c1=(sqr(a.y)-sqr(b.y)+sqr(a.x)-sqr(b.x))/2;
    double a2=c.x-a.x,b2=c.y-a.y,c2=(sqr(a.x)+sqr(a.y)-sqr(c.x)-sqr(c.y))/2;
    return tmp=cross(a1,a2,b1,b2,c1,c2);
}
int cmp(double x){
    if (fabs(x)<eps) return 0;
    if (x>0) return 1;
    return -1;
}
int main(){
    n=read();
    rep(i,1,n){
        scanf("%lf%lf",&a[i].x,&a[i].y);
    }
    int x,y;
    rep(i,1,n) {
        x=rand()%n+1; y=rand()%n+1;
        swap(a[x],a[y]);
    }
    o=a[1]; r=0;
    rep(i,2,n){
        if (cmp(dis(a[i],o)-r)<0) continue;
        o=a[i]; r=0;
        rep(j,1,i-1){
            if (cmp(dis(a[j],o)-r)<0) continue;
            o=get2(a[i],a[j]); r=dis(a[j],o);
            rep(k,1,j-1) {
                if (cmp(dis(a[k],o)-r)<0) continue;
                o=get3(a[i],a[j],a[k]); r=dis(a[k],o);
            }
        }
    }
    printf("%.2lf %.2lf ",o.x,o.y);
    printf("%.2lf\n",r);
    return 0; 
}