HDU 4773 Problem of Apollonius 圆的反演

传送门
坐标系上给定两个不相交的圆，求所有过一定点且与这两个圆都外切的圆。

反演变换

如果把以那个定点为原点进行反演变换，那么过定点的圆就变成了直线，不过定点的圆还是圆，问题就转化成了求那两个圆的公切线。求出所有公切线再反演回圆再判是否都是外切。

#include <bits/stdc++.h>
using namespace std;
#define prt(k) cerr<<#k" = "<<k<<endl
typedef long long LL;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
int cmp(double x)
{return x < -eps ? -1 : x > eps;
}
struct P
{double x, y, r;P() { x = 0, y = 0, r = 1; }P(double xx, double yy, double rr =  1){x = xx, y = yy, r = rr;}P getP(double a){return P(x + cos(a)*r, y + sin(a)*r);}void in() { scanf("%lf%lf%lf", &x, &y, &r); }P operator * (double k) { return P(x*k, y*k, r*k); }P operator / (double k) {assert(abs(k) > 1e-8);return P(x/k, y/k);}void out(){printf("%.10f %.10f %.10f\n", x, y, r);}
};
double pf(double x) { return x * x; }
P operator-(P a, P  b)
{return P(a.x - b.x, a.y - b.y);
}
P operator+(P a, P  b)
{return P(a.x + b.x, a.y + b.y);
}
double abs(P p) { return sqrt(pf(p.x) + pf(p.y)); }
double dis(P a, P b) { return abs(a - b); }
typedef P Circle ;
P a, b, o;
double dot(P a, P b) { return a.x*b.x + a.y*b.y; }
P cuizhu(P p, P s, P t) ///点到直线垂足
{double r = dot(t-s, p-s)/dot(t-s, t-s);return s + (t-s)*r;
}
const double r = 1;
/// 圆的反演（结果是一个圆），反演中心O为原点，反演满足OC * OC' = r*r  这里 r = 1
/// 圆不经过反演中心
Circle inv(Circle a)
{double d1  =  abs(a);double r1 = a.r;double r2 = r1 / (d1 * d1 - r1 * r1) * r * r;double d2 = (r1 * r2 + r * r) / d1;double x = a.x / d1 * d2;double y = a.y / d1 * d2;P p = a * (r2 / r1);p.r = r2;return P(x, y, r2);
}
Circle inv(P s, P t) /// 直线的反演，直线不经过反演中心
{P O(0,0);P p = cuizhu(O, s, t);double d = dis(O, p);p = p * (0.5 / d / d);p.r = dis(p, O);return p;
}
/// 圆的切线，返回切线条数，见白书 267 页
int getTan(Circle A, Circle B, P a[], P b[])
{int cnt = 0;if (A.r < B.r) { swap(A, B); swap(a, b); }double d2 = pf(dis(A, B));double rdiff = (A.r - B.r);double rsum = A.r + B.r;if (d2 < pf(rdiff)) return 0;double base = atan2(B.y - A.y, B.x - A.x);if (d2 == 0 && A.r == B.r)  return -1;if (d2 == pf(rdiff)) {a[cnt] = A.getP(base); b[cnt++] = B.getP(base);return cnt;}double ang = acos((A.r - B.r) / sqrt(d2));a[cnt] = A.getP(base+ang); b[cnt++] = B.getP(base + ang);a[cnt] = A.getP(base - ang); b[cnt++] = B.getP(base - ang);if (d2 == rsum * rsum) {a[cnt] = A.getP(base); b[cnt++] = B.getP(PI + base);}else if (d2 > rsum * rsum) {double ang = acos((A.r + B.r) / sqrt(d2));a[cnt] = A.getP(base + ang); b[cnt++] = B.getP(base + ang);a[cnt] = A.getP(base - ang); b[cnt++] = B.getP(base - ang);}return cnt;
}
bool XiangQie(P a, P b)
{return cmp(dis(a, b) - a.r - b.r) == 0;
}
bool in(Circle C, P p)
{return cmp(dis(C, p) - C.r) == 0;
}
Circle A, B;
bool check(Circle ans)
{return XiangQie(A, ans) && XiangQie(B, ans) && in(ans, o);
}
int main()
{int re; scanf("%d", &re);while (re--) {a.in(); b.in();A = a, B = b;scanf("%lf%lf", &o.x, &o.y);a.x -= o.x; a.y -= o.y;b.x -= o.x; b.y -= o.y;a = inv(a); b = inv(b);P p1[55], p2[55];int cnt = getTan(a, b, p1, p2);vector<Circle> v;for (int i=0;i<cnt;i++) {Circle ans = inv(p1[i], p2[i]);ans.x += o.x; ans.y += o.y;if (check(ans)) v.push_back(ans);}printf("%d\n", v.size());for (auto C : v) C.out();}
}

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