HDU 4773 圆的反演（经典

题目链接
参考博客

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题意：给定两个相离的圆和圆外定点P,求过点P且与已知的两个圆外切的所有圆。

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思路：
圆的反演变换入门题。
掌握两个关于圆反演的性质即可解决该问题。

一是相切两圆的反相仍相切，若反相均为直线，则平行。
二是经过反演中心的圆，其反相为直线。

故我们可以以P为反演中心，任意值为反演半径。先求出两个已知圆的反演圆，则与两个反演圆外切的直线，其反演后一定满足：
(1).与两个圆的原形相切。
(2).一定通过反演中心P点。

因题目是求外切，故注意一下反演圆圆心和P点应在切线同一侧，这样就能保证切线反演以后是与原形外切的圆。

代码：

#include<cmath>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const double eps = 1e-10;class Point{public:    double x,y;    Point(double _x=0,double _y=0):x(_x),y(_y){}    Point operator + (Point  rhs){return Point(x+rhs.x,y+rhs.y);}    Point operator - (Point  rhs){return Point(x-rhs.x,y-rhs.y);}    Point operator * (double rhs){return Point(x*rhs,y*rhs);}    Point operator / (double rhs){return Point(x/rhs,y/rhs);}    Point Move(double a,double d){return Point(x+d*cos(a),y+d*sin(a));}    void Read(){scanf("%lf%lf",&x,&y);}}P;class Circle{public:    Point o;double r;    Circle(double _x=0,double _y=0,double _r=0):o(_x,_y),r(_r){}    void Read(){o.Read();scanf("%lf",&r);}    void out(){printf("%.8f %.8f %.8f\n",o.x,o.y,r);}}c[5];int Sign(double x){return (x>eps) - (x<-eps);}double Cross(Point a,Point b,Point c){return (b.x-a.x)*(c.y-a.y) - (c.x-a.x)*(b.y-a.y);}double Dis(Point a,Point b){return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));}double R;int tot;Circle Inverse(Circle a){    Circle res;    double oc1 = Dis(P,a.o);    double k1 = 1.0 / (oc1 - a.r);    double k2 = 1.0 / (oc1 + a.r);    res.r = 0.5*(k1 - k2)*R*R;    double oc2 = 0.5*(k1 + k2)*R*R;    res.o =P + (a.o - P)*oc2 / oc1;    return res;}void Mark(Point a,Point b){    ++tot;    double tem = fabs(Cross(a,P,b)/Dis(a,b))*2.0;    c[tot].r = R*R/tem;    double d = Dis(a,c[0].o);    c[tot].o = P + (a-c[0].o)*(c[tot].r/d);}void solve(){    for(int i=0 ;i<2 ;i++) c[i] = Inverse(c[i]);    if(c[0].r < c[1].r) swap(c[0],c[1]);    Point tem = c[1].o - c[0].o;    double a1 = atan2(tem.y,tem.x);    double a2 = acos((c[0].r - c[1].r)/Dis(c[0].o,c[1].o));    Point P1 = c[0].o.Move(a1+a2,c[0].r);    Point P2 = c[1].o.Move(a1+a2,c[1].r);    if(Sign(Cross(P1,c[0].o,P2)) == Sign(Cross(P1,P,P2))) Mark(P1,P2);    P1 = c[0].o.Move(a1-a2,c[0].r);    P2 = c[1].o.Move(a1-a2,c[1].r);    if(Sign(Cross(P1,c[0].o,P2)) == Sign(Cross(P1,P,P2))) Mark(P1,P2);}int main(){    R = 5.0; //保证精度的前提下任意取的反演半径    int T;scanf("%d",&T);    while(T--){        tot = 1;        for(int i=0 ;i<2 ;i++) c[i].Read();        P.Read();        solve();        printf("%d\n",tot-1);        for(int i=2 ;i<=tot ;i++) c[i].out();    }    return 0;}