题意：在二维平面上求解6个子问题：
1.三角形的外接圆；
2.三角形的内切圆；
3.点到圆的切线；
4.过定点且与定直线相切的半径为r的圆；
5.同时与两相交直线相切的半径为r的圆；
6.同时与两相离圆相切的半径为r的圆。
题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=24008
——>>做到第4小问的时候，出现了一个精度问题：本来用dcmp(delta) == 0来判断直线与圆相切，可以在此处硬是不正确。。。书中有言：“圆和直线相切的判定很容易受到误差影响，因为这里的直线是计算出来的，而不是题目中输入的。”。。。所以——“特殊判断圆心到输入直线的距离”（“当然也可以通过调整eps的数值来允许一定的误差，但不推荐这样做。调节eps只是掩盖了问题，并没有消除“）。
代码虽长，但1A啦。。。
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>using namespace std;const int maxn = 50;
const double eps = 1e-10; //调到1e-6以上第4问就可以用delta判断切线,但《训练指南》建议，尽量不要调eps
const double pi = acos(-1);char type[maxn];int dcmp(double x){return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1);
}struct Point{double x;double y;Point(double x = 0, double y = 0):x(x), y(y){}bool operator < (const Point& e) const{return dcmp(x - e.x) < 0 || (dcmp(x - e.x) == 0 && dcmp(y - e.y) < 0);}bool operator == (const Point& e) const{return dcmp(x - e.x) == 0 && dcmp(y - e.y) == 0;}int read(){return scanf("%lf%lf", &x, &y);}
}p[3];typedef Point Vector;Vector operator + (Point A, Point B){return Vector(A.x + B.x, A.y + B.y);
}Vector operator - (Point A, Point B){return Vector(A.x - B.x, A.y - B.y);
}Vector operator * (Point A, double p){return Vector(A.x * p, A.y * p);
}Vector operator / (Point A, double p){return Vector(A.x / p, A.y / p);
}struct Line{Point p;Point v;Line(){}Line(Point p, Point v):p(p), v(v){}int read(){return scanf("%lf%lf%lf%lf", &p.x, &p.y, &v.x, &v.y);}Point point(double t){return p + v * t;}
};struct Circle{Point c;double r;Circle(){}Circle(Point c, double r):c(c), r(r){}int read(){return scanf("%lf%lf%lf", &c.x, &c.y, &r);}Point point(double a){return Point(c.x + r * cos(a), c.y + r * sin(a));}
};double Dot(Vector A, Vector B){return A.x * B.x + A.y * B.y;
}double Cross(Vector A, Vector B){return A.x * B.y - B.x * A.y;
}double Length(Vector A){return sqrt(Dot(A, A));
}Vector Rotate(Vector A, double rad){return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}Vector Normal(Vector A){double L = Length(A);return Vector(-A.y / L, A.x / L);
}double DistanceToLine(Point P, Point A, Point B){ //点到直线的距离Vector v1 = B - A;Vector v2 = P - A;return fabs(Cross(v1, v2) / Length(v1));
}double angle(Vector v){ //求向量的极角return atan2(v.y, v.x);
}Point GetLineIntersection(Line l1, Line l2){ //求两直线的交点（前提：相交）Vector u = l1.p - l2.p;double t = Cross(l2.v, u) / Cross(l1.v, l2.v);return l1.point(t);
}int getLineCircleIntersection(Line l, Circle C, double& t1, double& t2, vector<Point>& sol){ //求直线与圆的交点double a = l.v.x;double b = l.p.x - C.c.x;double c = l.v.y;double d = l.p.y - C.c.y;double e = a * a + c * c;double f = 2 * (a * b + c * d);double g = b * b + d * d - C.r * C.r;double delta = f * f - 4 * e * g;double dist = DistanceToLine(C.c, l.p, l.p+l.v);if(dcmp(dist - C.r) == 0){ //相切，此处需特殊判断，不能用deltat1 = t2 = -f / (2 * e);sol.push_back(l.point(t1));return 1;}if(dcmp(delta) < 0) return 0; //相离else{ //相交t1 = (-f - sqrt(delta)) / (2 * e); sol.push_back(l.point(t1));t2 = (-f + sqrt(delta)) / (2 * e); sol.push_back(l.point(t2));return 2;}
}int GetCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol){ //求圆与圆的交点double d = Length(C1.c - C2.c);if(dcmp(d) == 0){if(dcmp(C1.r - C2.r) == 0) return -1; //两圆重合return 0; //同心圆但不重合}if(dcmp(C1.r + C2.r - d) < 0) return 0; //外离if(dcmp(fabs(C1.r - C2.r) - d) > 0) return 0; //内含double a = angle(C2.c - C1.c);double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d));Point p1 = C1.point(a + da);Point p2 = C1.point(a - da);sol.push_back(p1);if(p1 == p2) return 1; //外切sol.push_back(p2);return 2;
}Circle CircumscribedCircle(Point p1, Point p2, Point p3){ //求三角形的外心double Bx = p2.x - p1.x, By = p2.y - p1.y;double Cx = p3.x - p1.x, Cy = p3.y - p1.y;double D = 2 * (Bx * Cy - By * Cx);double cx = (Cy * (Bx * Bx + By * By) - By * (Cx * Cx + Cy * Cy)) / D + p1.x;double cy = (Bx * (Cx * Cx + Cy * Cy) - Cx * (Bx * Bx + By * By)) / D + p1.y;Point p(cx, cy);return Circle(p, Length(p1-p));
}Circle InscribedCircle(Point p1, Point p2, Point p3){ //求三角形的内切圆double a = Length(p3 - p2);double b = Length(p3 - p1);double c = Length(p2 - p1);Point p = (p1 * a + p2 * b + p3 * c) / (a + b + c);return Circle(p, DistanceToLine(p, p2, p3));
}int TangentLineThroughPoint(Point p, Circle C, Vector *v){ //求点到圆的直线Vector u = C.c - p;double dist = Length(u);if(dcmp(dist - C.r) < 0) return 0;else if(dcmp(dist - C.r) < eps){v[0] = Rotate(u, pi / 2);return 1;}else{double ang = asin(C.r / dist);v[0] = Rotate(u, ang);v[1] = Rotate(u, -ang);return 2;}
}void CircleThroughAPointAndTangentToALineWithRadius(Point p, Point p1, Point p2, double r){Vector AB = p2 - p1;Vector change1 = Rotate(AB, pi / 2) / Length(AB) * r;Vector change2 = Rotate(AB, -pi / 2) / Length(AB) * r;Line l1(p1 + change1, AB);Line l2(p1 + change2, AB);vector<Point> sol;sol.clear();double t1, t2;int cnt1 = getLineCircleIntersection(l1, Circle(p, r), t1, t2, sol);int cnt2 = getLineCircleIntersection(l2, Circle(p, r), t1, t2, sol);int cnt = cnt1 + cnt2;if(cnt) sort(sol.begin(), sol.end());printf("[");for(int i = 0; i < cnt; i++){printf("(%.6f,%.6f)", sol[i].x, sol[i].y);if(cnt == 2 && !i) printf(",");}puts("]");
}void CircleTangentToTwoLinesWithRadius(Point A, Point B, Point C, Point D, double r){Vector AB = B - A;Vector change = Normal(AB) * r;Point newA1 = A + change;Point newA2 = A - change;Vector CD = D - C;Vector update = Normal(CD) * r;Point newC1 = C + update;Point newC2 = C - update;Point p[5];p[0] = GetLineIntersection(Line(newA1, AB), Line(newC1, CD));p[1] = GetLineIntersection(Line(newA1, AB), Line(newC2, CD));p[2] = GetLineIntersection(Line(newA2, AB), Line(newC1, CD));p[3] = GetLineIntersection(Line(newA2, AB), Line(newC2, CD));sort(p, p + 4);printf("[");printf("(%.6f,%.6f)", p[0].x, p[0].y);for(int i = 1; i < 4; i++){printf(",(%.6f,%.6f)", p[i].x, p[i].y);}puts("]");
}void CircleTangentToTwoDisjointCirclesWithRadius(Circle C1, Circle C2, double r){Vector CC = C2.c - C1.c;double rdist = Length(CC);if(dcmp(2 * r - rdist + C1.r + C2.r) < 0) puts("[]");else if(dcmp(2 * r - rdist + C1.r + C2.r) == 0){double ang = angle(CC);Point A = C1.point(ang);Point B = C2.point(ang + pi);Point ret = (A + B) / 2;printf("[(%.6f,%.6f)]\n", ret.x, ret.y);}else{Circle A = Circle(C1.c, C1.r + r);Circle B = Circle(C2.c, C2.r + r);vector<Point> sol;sol.clear();GetCircleCircleIntersection(A, B, sol);sort(sol.begin(), sol.end());printf("[(%.6f,%.6f),(%.6f,%.6f)]\n", sol[0].x, sol[0].y, sol[1].x, sol[1].y);}
}int main()
{while(scanf("%s", type) == 1){if(strcmp(type, "CircumscribedCircle") == 0){Point p1, p2, p3;p1.read();p2.read();p3.read();Circle ret = CircumscribedCircle(p1, p2, p3);printf("(%f,%f,%f)\n", ret.c.x, ret.c.y, ret.r);}else if(strcmp(type, "InscribedCircle") == 0){Point p1, p2, p3;p1.read();p2.read();p3.read();Circle ret = InscribedCircle(p1, p2, p3);printf("(%f,%f,%f)\n", ret.c.x, ret.c.y, ret.r);}else if(strcmp(type, "TangentLineThroughPoint") == 0){Circle C;Point p;C.read();p.read();Vector v[3];int cnt = TangentLineThroughPoint(p, C, v);double ret[3];for(int i = 0; i < cnt; i++){ret[i] = angle(v[i]);if(dcmp(ret[i] - pi) == 0) ret[i] = 0;if(dcmp(ret[i]) < 0) ret[i] += pi;}sort(ret, ret + cnt);printf("[");for(int i = 0; i < cnt; i++){printf("%.6f", ret[i] / pi * 180);if(cnt == 2 && !i) printf(",");}puts("]");}else if(strcmp(type, "CircleThroughAPointAndTangentToALineWithRadius") == 0){Point p, p1, p2;double r;p.read();p1.read();p2.read();scanf("%lf", &r);CircleThroughAPointAndTangentToALineWithRadius(p, p1, p2, r);}else if(strcmp(type, "CircleTangentToTwoLinesWithRadius") == 0){Point A, B, C, D;double r;A.read();B.read();C.read();D.read();scanf("%lf", &r);CircleTangentToTwoLinesWithRadius(A, B, C, D, r);}else{Circle C1, C2;double r;C1.read();C2.read();scanf("%lf", &r);CircleTangentToTwoDisjointCirclesWithRadius(C1, C2, r);}}return 0;
}