# 最小圆覆盖

 1 struct Point{double x,y;};
2 struct Circle{Point c;double r;};
3 double dist(Point a,Point b){
4     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
5 }
6 Circle calc(Point p1,Point p2,Point p3){
7     Circle temp;
8     double a,b,c,d,e,f;
9     a=p2.x-p1.x;
10     b=p2.y-p1.y;
11     c=(p2.x*p2.x+p2.y*p2.y-p1.x*p1.x-p1.y*p1.y)/2;
12     d=p3.x-p1.x;
13     e=p3.y-p1.y;
14     f=(p3.x*p3.x+p3.y*p3.y-p1.x*p1.x-p1.y*p1.y)/2;
15     temp.c.y=(c*d-f*a)/(b*d-e*a);
16     temp.c.x=(c*e-f*b)/(a*e-b*d);
17     return temp;
18 }
19 Circle minC(Point *p,int n){
20     Circle O;
21     int i,j,k;
22     O.c=p[0];O.r=0;
23     for(i=1;i<n;i++){
24         if(dist(O.c,p[i])<=O.r+1e-6)continue;
25         O.c=p[i];O.r=0;
26         for(j=0;j<i;j++){
27             if(dist(O.c,p[j])<=O.r+1e-6)continue;
28             O.c.x=(p[i].x+p[j].x)/2;O.c.y=(p[i].y+p[j].y)/2;O.r=dist(O.c,p[j]);
29             for(k=0;k<j;k++){
30                 if(dist(O.c,p[k])<=O.r+1e-6)continue;
31                 O=calc(p[i],p[j],p[k]);
32                 O.r=dist(O.c,p[k]);
33             }
34         }
35     }
36     return O;
37 }    

posted @ 2017-10-15 15:22  weeping  阅读(202)  评论(0编辑  收藏  举报