bzoj 1337 最小圆覆盖

 /**************************************************************
     Problem: 1337
     User: idy002
     Language: C++
     Result: Accepted
     Time:4 ms
     Memory:2372 kb
 ****************************************************************/

 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #define line(a,b) ((b)-(a))
 #define eps 1e-10
 #define N 100010
 using namespace std;

 int sg( double x ) { return (x>-eps)-(x<eps); }
 struct Vector {
     double x, y;
     Vector( double x=, double y= ):x(x),y(y){}
     Vector operator+( const Vector &b ) const { return Vector(x+b.x,y+b.y); }
     Vector operator-( const Vector &b ) const { return Vector(x-b.x,y-b.y); }
     Vector operator*( double b ) const { return Vector(x*b,y*b); }
     Vector operator/( double b ) const { return Vector(x/b,y/b); }
     double operator^( const Vector &b ) const { return x*b.y-y*b.x; }
     double len() { return sqrt(x*x+y*y); }
     Vector nor() { return Vector(-y,x); }
 };
 typedef Vector Point;
 Point inter( Point p, Vector u, Point q, Vector v ) {
     return p+u*((line(p,q)^v)/(u^v));
 }
 struct Circle {
     Point o;
     double r;
     Circle(){}
     Circle( Point a ):o(a),r(){}
     Circle( Point a, Point b ) {
         o = (a+b)/;
         r = (a-b).len()/;
     }
     Circle( Point a, Point b, Point c ) {   //  ab^bc != 0
         Point p=(a+b)/, q=(b+c)/;
         Vector u=(a-b).nor(), v=(b-c).nor();
         o = inter(p,u,q,v);
         r = (o-a).len();
     }
     bool contain( Point a ) {
         return sg( (a-o).len() - r ) <= ;
     }
 };

 int n;
 Point pts[N];

 int main() {
     scanf( "%d", &n );
     for( int i=; i<=n; i++ ) {
         double x, y;
         scanf( "%lf%lf", &x, &y );
         pts[i] = Point(x,y);
     }
     random_shuffle( pts+, pts++n );
     Circle c = Circle(pts[]);
     for( int i=; i<=n; i++ ) {
         if( c.contain(pts[i]) ) continue;
         c = Circle(pts[i]);
         for( int j=; j<i; j++ ) {
             if( c.contain(pts[j]) ) continue;
             c = Circle(pts[i],pts[j]);
             for( int k=; k<j; k++ ) {
                 if( c.contain(pts[k]) ) continue;
                 c = Circle(pts[i],pts[j],pts[k]);
             }
         }
     }
     printf( "%.3lf\n", c.r );
 }

题解见bzoj 1336