Bzoj 1336&1337 Alien最小圆覆盖

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cmath>
 5 using namespace std;
 6 const double eps=1e-8;
 7 const int mxn=100000;
 8 int n;
 9 struct point{
10     double x,y;
11     friend point operator +(const point a,const point b){
12         return (point){a.x+b.x,a.y+b.y};
13     }
14     friend point operator -(const point a,const point b){
15         return (point){a.x-b.x,a.y-b.y};
16     }
17     friend point operator /(const point a,double b){
18         return (point){a.x/b,a.y/b};
19     }
20 }p[mxn];
21 inline double dis(point a,point b){
22     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
23 }
24 
25 point center(point a,point b,point c){//返回三角形外心 
26     double a1,a2,b1,b2,c1,c2;
27     point ans;
28     a1=2*(b.x-a.x);b1=2*(b.y-a.y);c1=(b.x*b.x)-(a.x*a.x)+(b.y*b.y)-(a.y*a.y);
29     //c1=(a1*a1+b1*b1)/2
30     a2=2*(c.x-a.x);b2=2*(c.y-a.y);c2=(c.x*c.x)-(a.x*a.x)+(c.y*c.y)-(a.y*a.y);
31     //c2=(a2*a2+b2*b2)/2
32     if(fabs(a1)<eps){
33         ans.y=c1/b1;
34         ans.x=(c2-ans.y*b2)/a2;
35     }
36     else if(fabs(b1)<eps){
37         ans.x=c1/a1;
38         ans.y=(c2-ans.x*a2)/b2;
39     }
40     else{
41         ans.x=(c2*b1-c1*b2)/(a2*b1-a1*b2);
42         ans.y=(c2*a1-c1*a2)/(b2*a1-b1*a2);
43     }
44     return ans;
45 }
46 int main(){
47     scanf("%d",&n);
48     int i,j,k;
49     for(i=1;i<=n;i++){
50         scanf("%lf%lf",&p[i].x,&p[i].y);
51     }
52     random_shuffle(p+1,p+n+1);
53     point t=p[1];
54     double r=0.0;
55     for(i=2;i<=n;i++)//
56       if(dis(t,p[i])>r+eps){
57           t=(p[i]+p[1])/2;//默认圆心，等待增量 
58           r=dis(p[i],t);//半径 
59           for(j=2;j<i;j++)// 
60             if(dis(t,p[j])>r+eps){//若有点在圆外，更新圆心 
61                 t=(p[i]+p[j])/2;
62                 r=dis(t,p[i]);
63                 for(k=1;k<j;k++){//最多三点确定一圆 
64                     if(dis(p[k],t)>r+eps){
65                         t=center(p[i],p[j],p[k]);
66                         r=dis(p[i],t);
67                 }
68             }
69         }          
70     }
71     printf("%.10lf\n%.10lf %.10lf",r,t.x,t.y);
72     return 0;
73 }