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.