是cf933C的升级版。

平面图欧拉定理。over！

f=e-v+c+1

c是联通块，相交才视为一块。

e是圆弧数，v是顶点数。

#include 
    
     #define pii pair
     
      #define mp make_pair#define fi first#define se second#define pb push_backusing namespace std;typedef double db;const db eps=1e-6;const db pi=acos(-1);int sign(db k){    if (k>eps) return 1; else if (k<-eps) return -1; return 0;}int cmp(db k1,db k2){return sign(k1-k2);}int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内struct point{    db x,y;    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}    point operator * (db k1) const{return (point){x*k1,y*k1};}    point operator / (db k1) const{return (point){x/k1,y/k1};}    int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}    // 逆时针旋转    point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}    point turn90(){return (point){-y,x};}    bool operator < (const point k1) const{        int a=cmp(x,k1.x);        if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;    }    db abs(){return sqrt(x*x+y*y);}    db abs2(){return x*x+y*y;}    db dis(point k1){return ((*this)-k1).abs();}    point unit(){db w=abs(); return (point){x/w,y/w};}    void scan(){double k1,k2; scanf("%lf%lf",&k1,&k2); x=k1; y=k2;}    void print(){printf("%.11lf %.11lf\n",x,y);}    db getw(){return atan2(y,x);}    point getdel(){if (sign(x)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);}    int getP() const{return sign(y)==1||(sign(y)==0&&sign(x)==-1);}};int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}db rad(point k1,point k2){return atan2(cross(k1,k2),dot(k1,k2));}// -pi -> piint compareangle (point k1,point k2){//极角排序+    return k1.getP()
      
       0);}point proj(point k1,point k2,point q){ // q 到直线 k1,k2 的投影    point k=k2-k1;return k1+k*(dot(q-k1,k)/k.abs2());}point reflect(point k1,point k2,point q){return proj(k1,k2,q)*2-q;}int clockwise(point k1,point k2,point k3){// k1 k2 k3 逆时针 1 顺时针 -1 否则 0    return sign(cross(k2-k1,k3-k1));}int checkLL(point k1,point k2,point k3,point k4){// 求直线 (L) 线段 (S)k1,k2 和 k3,k4 的交点    return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))!=0;}point getLL(point k1,point k2,point k3,point k4){    db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);}int intersect(db l1,db r1,db l2,db r2){    if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1;}int checkSS(point k1,point k2,point k3,point k4){    return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&           sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&&           sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;}db disSP(point k1,point k2,point q){    point k3=proj(k1,k2,q);    if (inmid(k1,k2,k3)) return q.dis(k3); else return min(q.dis(k1),q.dis(k2));}db disSS(point k1,point k2,point k3,point k4){    if (checkSS(k1,k2,k3,k4)) return 0;    else return min(min(disSP(k1,k2,k3),disSP(k1,k2,k4)),min(disSP(k3,k4,k1),disSP(k3,k4,k2)));}int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;}struct circle{    point o; db r;    void scan(){o.scan(); scanf("%lf",&r);}    int inside(point k){return cmp(r,o.dis(k));}};int checkposCC(circle k1,circle k2){// 返回两个圆的公切线数量    if (cmp(k1.r,k2.r)==-1) swap(k1,k2);    db dis=k1.o.dis(k2.o);  int w1=cmp(dis,k1.r+k2.r),w2=cmp(dis,k1.r-k2.r);    if (w1>0) return 4; else if (w1==0) return 3; else if (w2>0) return 2;    else if (w2==0) return 1; else return 0;}vector
       
         getCC(circle k1,circle k2){// 沿圆 k1 逆时针给出 , 相切给出两个    int pd=checkposCC(k1,k2); if (pd==0||pd==4) return {};    db a=(k2.o-k1.o).abs2(),cosA=(k1.r*k1.r+a-k2.r*k2.r)/(2*k1.r*sqrt(max(a,(db)0.0)));    db b=k1.r*cosA,c=sqrt(max((db)0.0,k1.r*k1.r-b*b));    point k=(k2.o-k1.o).unit(),m=k1.o+k*b,del=k.turn90()*c;    return {m-del,m+del};}int t,n,vis[55];circle c[55];set
        
          s;vector
         
           v;set
          
            st;void slove(int x){ vis[x]=1; for(auto p:st){ for(int i=0;i