HDU

题目链接点这里

圆的反演主要有3条性质

//1.不过反演中心的圆经过反演变换仍然是一个不过反演中心的圆.

//2.不过反演中心的直线经过反演变换是一个经过反演中心的圆.

//3.反演变换不改变图形的相切性.

然后这道题就解决了。。

//注意精度，直接用rad计算圆弧中点，#include<iostream>#include<cstdio>#include<math.h>#include<algorithm>#include<map>#include<set>#include<bitset>#include<stack>#include<queue>#include<string.h>#include<string>#include<cstring>#include<vector>#include<time.h>#include<stdlib.h>using namespace std;#define INF 0x3f3f3f3f#define INFLL 0x3f3f3f3f3f3f3f3f#define FIN freopen("input.txt","r",stdin);#define mem(x,y) memset(x,y,sizeof(x));typedef unsigned long long ULL;typedef long long LL;#define fuck(x) cout<<x<<endl;const int  MX=333;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef pair<pair<int,int>,int> PIII;typedef pair<int,int> PII;const double eps=1e-8;const double PI=acos(-1);struct Point {    double x, y;    Point() {}    Point(double x,double y):x(x),y(y) {}};typedef Point Vector;int dcmp(double x) { //返回x的正负    if(fabs(x)<eps)return 0;    return x<0?-1:1;}Vector operator-(Vector A,Vector B) {    return Vector(A.x - B.x, A.y - B.y);}Vector operator+(Vector A,Vector B) {    return Vector(A.x + B.x, A.y + B.y);}Vector operator*(Vector A,double p) {    return Vector(A.x*p, A.y*p);}Vector operator/(Vector A,double p) {    return Vector(A.x/p, A.y/p);}bool operator<(const Point&a,const Point&b) {    return a.x<b.x||(a.x==b.x&&a.y<b.y);}bool operator==(const Point&a,const Point&b) {    return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}double Dot(Vector A,Vector B) { //点积    return A.x*B.x+A.y*B.y;//如果改成整形记得加LL}double Cross(Vector A,Vector B) { //叉积    return A.x*B.y-A.y*B.x;//如果改成整形记得加LL}//向量长度double Length(Vector A) {    return sqrt(Dot(A,A));}//2个向量之间的夹角double Angle(Vector A,Vector B) {    return acos(Dot(A,B)/Length(A)/Length(B));}//向量的极角double angle(Vector v) {    return atan2(v.y,v.x);}//将A向量逆时针旋转radVector Rotate( Vector A,double rad) {    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}//返回A的逆时针旋转90度的单位法向量Vector Normal(Vector A) {    double L=Length(A);    return Vector(-A.y/L,A.x/L);}//计算2条直线P+tv和Q+tw的交点，请先确保不是平行(v！=w)Point GetLineIntersection(Point P,Vector v,Point Q,Vector w) {    Vector u=P-Q;    double t=Cross(w,u)/Cross(v,w);    return P+v*t;}//P到直线AB的距离double DistanceToLine(Point P,Point A,Point B) {    Vector v1=B-A,v2=P-A;    return fabs(Cross(v1,v2))/Length(v1);}//园struct Circle {    Point c;    double r;    Circle() {}    Circle(Point c,double r):c(c),r(r) {}    Point getpoint(double rad) {        return Point(c.x+cos(rad)*r,c.y+sin(rad)*r);    }};//有向线段struct Line {    Point p;    Vector v;//方向向量，左边为半平面    double ang;//极角    Line() {}    Line(Point p,Vector v):p(p),v(v) {        ang=atan2(v.y,v.x);    }    bool operator<(const Line &L)const { //极角排序        return ang<L.ang;    }    Point point(double t) {        return p + v*t;    }    Line move(double d) { //向左边平移d单位        return Line(p + Normal(v)*d, v);    }};//返回切线条数-1表示无穷多条切线//a[i],b[i]分别是第i条切线在圆A，圆B上的切点int getCircleTangents(Circle A,Circle B,Point *a,Point *b) {    int cnt=0;//切线条数    if(A.r<B.r) {        swap(A,B);        swap(a,b);    }    double d2=(A.c.x-B.c.x)*(A.c.x-B.c.x)+(A.c.y-B.c.y)*(A.c.y-B.c.y);//圆心距    double rdiff=A.r-B.r;    double rsum=A.r+B.r;    if(dcmp(d2-rdiff*rdiff)<0) return 0;//内含    double base=atan2(B.c.y-A.c.y,B.c.x-A.c.x);//求出圆心连线的极角    if(dcmp(d2)==0&&dcmp(A.r-B.r)==0) return -1;//两圆重合    if(dcmp(d2-rdiff*rdiff)==0) {  //内切 一条外公切线        a[cnt]=A.getpoint(base);        b[cnt]=B.getpoint(base);        cnt++;        return 1;    }    //有两条外公切线    double ang=acos((A.r-B.r)/sqrt(d2));//求出切线与圆心连线的夹角    a[cnt]=A.getpoint(base+ang);    b[cnt]=B.getpoint(base+ang);    cnt++;    a[cnt]=A.getpoint(base-ang);    b[cnt]=B.getpoint(base-ang);    cnt++;    if(d2==rsum*rsum) { //两圆外切        a[cnt]=A.getpoint(base);        b[cnt]=B.getpoint(base+PI);        cnt++;    } else if(d2>rsum*rsum) {        double ang=acos((A.r+B.r)/sqrt(d2));//求出内公切线和圆心连线的夹角        a[cnt]=A.getpoint(base+ang);        b[cnt]=B.getpoint(PI+base+ang);        cnt++;        a[cnt]=A.getpoint(base-ang);        b[cnt]=B.getpoint(PI+base-ang);        cnt++;    }    return cnt;}//求圆u关于c的反演圆Circle InvCircletoCircle(Circle C,Circle u) {    Circle T;    double t = Length(C.c-u.c);    double x = 1.0 / (t - u.r);    double y = 1.0 / (t + u.r);    T.r = (x - y)*C.r*C.r/ 2.0;    double s = (x + y)*C.r*C.r/ 2.0;    T.c = C.c + (u.c - C.c) * (s / t);    return T;}//求直线u关于c的反演圆Circle InvLinetoCircle(Circle C,Line u) {    Circle T;    Point w=GetLineIntersection(C.c,Normal(u.v),u.p,u.v);//垂足    double dis = Length(w-C.c);    double t = C.r*C.r/ dis;    Point p=C.c+(w-C.c)/dis*t;    T.r=t/2;    T.c=(p+C.c)/2;    return T;}//求圆u关于c的反演直线//先保证u经过C的圆心Line InvCircletoLine(Circle C,Circle u) {    Line T;    double t=C.r*C.r/(2*u.r);    Point p=(u.c-C.c)/Length(u.c-C.c)*t+C.c;//垂足    T.p=p;    T.v=Normal(u.c-C.c);    return T;}int main() {    //FIN;    int T;    cin>>T;    while(T--) {        Circle a,b,inva,invb,C;        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&a.c.x,&a.c.y,&a.r,&b.c.x,&b.c.y,&b.r,&C.c.x,&C.c.y);        C.r=3;        inva=InvCircletoCircle(C,a);        invb=InvCircletoCircle(C,b);        //cout<<inva.c.x<<" "<<inva.c.y<<" "<<inva.r<<endl;        //cout<<invb.c.x<<" "<<invb.c.y<<" "<<invb.r<<endl;        Point pa[10],pb[10];        int t=getCircleTangents(inva,invb,pa,pb);        vector<Circle> ans;        //cout<<t<<endl;        for(int i=0; i<t; i++) {            int t1=dcmp(Cross(pa[i]-pb[i],inva.c-pb[i]));            int t2=dcmp(Cross(pa[i]-pb[i],invb.c-pb[i]));            int t3=dcmp(Cross(pa[i]-pb[i],C.c-pb[i]));            //cout<<t1<<t2<<t3<<endl;            if(t1==t2&&t2==t3) {                ans.push_back(InvLinetoCircle(C,Line(pa[i],pa[i]-pb[i])));            }        }        cout<<ans.size()<<endl;        for(auto i:ans) {            printf("%.8f %.8f %.8f\n",i.c.x,i.c.y,i.r);        }    }    return 0;}