bzoj2823

最小圆覆盖

有个东西叫作随机增量法，具体可以baidu

这里来说说怎么求三点共圆

这其实就是求两条线段的交点

在编程中，我们解方程是比较麻烦的一个比较好的方法是利用相似三角形

设线段AB，CD交P，则PC：PD=Sabc:Sabd

然后用定比分点就可以求的交点坐标了

 const eps=1e-6;

 type point=record
        x,y:double;
      end;

 var p:array[..] of point;
     i,n,j,k:longint;
     r:double;

 procedure swap(var a,b:point);
   var c:point;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 function dis(i,j:longint):double;
   begin
     exit(sqrt(sqr(p[i].x-p[j].x)+sqr(p[i].y-p[j].y)));
   end;

 function cov(i:longint):boolean;
   begin
     if dis(i,)-r>eps then exit(false)
     else exit(true);
   end;

 procedure little(i,j:longint);
   begin
     p[].x:=(p[i].x+p[j].x)/;
     p[].y:=(p[i].y+p[j].y)/;
     r:=dis(i,j)/;
   end;

 function cross(x1,y1,x2,y2:double):double;
   begin
     exit(x1*y2-x2*y1);
   end;

 procedure circle(i,j,k:longint);
   var ret,xa,ya,xb,yb,xc,yc,xd,yd,t1,t2:double;
   begin
     ret:=/dis(i,j);
     xa:=(p[i].x+p[j].x)/; ya:=(p[i].y+p[j].y)/;
     xb:=xa-ret*(p[i].y-p[j].y); yb:=ya+ret*(p[i].x-p[j].x);
     ret:=/dis(j,k);
     xc:=(p[j].x+p[k].x)/; yc:=(p[j].y+p[k].y)/;
     xd:=xc-ret*(p[j].y-p[k].y); yd:=yc+ret*(p[j].x-p[k].x);
     t1:=cross(xc-xa,yc-ya,xb-xa,yb-ya);
     t2:=cross(xb-xa,yb-ya,xd-xa,yd-ya);
     p[].x:=(t1*xd+t2*xc)/(t1+t2);
     p[].y:=(t1*yd+t2*yc)/(t1+t2);
     r:=dis(,i);
  end;

 begin
   readln(n);
   for i:= to n do
   begin
     readln(p[i].x,p[i].y);
     swap(p[i],p[trunc(random*i)+]);
   end;
   little(,);
   for i:= to n do
     if not cov(i) then
     begin
       little(,i);
       for j:= to i- do
         if not cov(j) then
         begin
           little(i,j);
           for k:= to j- do
             if not cov(k) then circle(i,j,k);
         end;
     end;
   writeln(p[].x::,' ',p[].y::,' ',r::);
 end.