bzoj1336: [Balkan2002]Alien最小圆覆盖

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1336

1336: [Balkan2002]Alien最小圆覆盖

Time Limit: 1 Sec  Memory Limit: 162 MBSec  Special Judge
Submit: 1608  Solved: 713
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Description

给出N个点，让你画一个最小的包含所有点的圆。

Input

先给出点的个数N,2<=N<=100000，再给出坐标Xi,Yi.(-10000.0<=xi,yi<=10000.0)

Output

输出圆的半径，及圆心的坐标

Sample Input

6
8.0 9.0
4.0 7.5
1.0 2.0
5.1 8.7
9.0 2.0
4.5 1.0

Sample Output

5.00
5.00 5.00

HINT

 

Source

最小圆覆盖裸题。

至于最小圆覆盖怎么写，看这篇博客吧，挺详细的http://blog.sina.com.cn/s/blog_6e63f59e010120dl.html

好久没写博客了……

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define maxn 100005
using namespace std;
int n;
double r;
const double eps=1e-9;
struct fuck{double x,y;}p[maxn],o;
fuck operator +(fuck x,fuck y){return (fuck){x.x+y.x,x.y+y.y};}
fuck operator -(fuck x,fuck y){return (fuck){x.x-y.x,x.y-y.y};}
fuck operator *(fuck x,double y){return (fuck){x.x*y,x.y*y};}
fuck operator /(fuck x,double y){return (fuck){x.x/y,x.y/y};}
double sqr(double x){return x*x;}
double dis(fuck x,fuck y){return sqr(x.x-y.x)+sqr(x.y-y.y);}
void get(fuck x,fuck y,double &a,double &b,double &c){
    a=2*(y.x-x.x);b=2*(y.y-x.y);c=(sqr(x.x)+sqr(x.y)-sqr(y.x)-sqr(y.y));
}
void calc(double a,double b,double c,double d,double e,double f){
    o.y=(a*f-c*d)/(b*d-e*a);
    o.x=(b*f-c*e)/(a*e-b*d);
}
int main(){
    srand(1315313);
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
    random_shuffle(p+1,p+n+1);
    o=p[1];r=0;
    for(int i=2;i<=n;i++)if(dis(o,p[i])>r){
            o=p[i];r=0;
            for(int j=1;j<i;j++)if(dis(p[j],o)>r){
                    o=(p[j]+p[i])/2;r=dis(p[j],o);
                    for(int k=1;k<j;k++)if(dis(p[k],o)>r){
                        double a,b,c,d,e,f;
                        get(p[i],p[j],a,b,c);get(p[j],p[k],d,e,f);
                        calc(a,b,c,d,e,f);r=dis(o,p[k]);
                    }
                }
        }
    printf("%f
",sqrt(r));
    printf("%f %f
",o.x,o.y);
}