10 node p[200000]; 11 double r;7 W2 x9 ~ K9 X
12 node O;
+ u* [# X. e( |7 J) W" @! [ 13 double dist(node a,node b)+ r6 p; C, `+ U8 x4 O
14 {' | D. ?9 k8 L. Q8 ]5 K2 v, m
15 return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );
3 N8 Z, r. P, W5 W$ U5 W9 I9 Z 16 }
v4 Y- o, r" `6 ]; p9 w 17 void calc(double a,double b,double c,double d,double e,double f) //给出两条直线ax+by+c=0,dx+ey+f=0 求交点* r6 }& B2 m4 z+ l6 O
18 { //注意到三角形里两条中垂线不可能平行,所以不会产生除0错误% J3 ^% p8 V* ` o
19 O.y=(c*d-f*a)/(b*d-e*a);
# O* }6 K2 A/ b" M s 20 O.x=(c*e-f*b)/(a*e-b*d);$ w$ P: u0 [6 m6 ?* O) M
21 }
q$ Y2 M) f3 ?' W: Q) Z% Q 22 int main()
1 R M$ Z2 U* E4 b$ o% X( W4 F% f# ? 23 {0 A8 y |0 w6 A
24 freopen("HYOJ1337.in","r",stdin);& v& c% h6 L( C5 S1 A6 L2 J1 ~
25 freopen("HYOJ1337.out","w",stdout);' L/ r& Z s; h8 B
26 scanf("%d",&n);* P0 D9 N$ t$ l! V- o. {
$ ? `3 S+ `4 j1 C! `7 y* K d& k
27 for (int i=1;i |