UVa12304

基础题，注意精度和旋转方向。

 

#include <iostream>

#include <math.h>

#include <vector>

#include <algorithm>

#include <string>



using namespace std;



#define PI acos(-1.0)



#define M 100007

#define N 65736



const int inf = 0x7f7f7f7f;

const int mod = 1000000007;

const double eps = 1e-6;





struct Point

{

	double x, y;

	Point(double tx = 0, double ty = 0) : x(tx), y(ty){}

};

typedef Point Vtor;

//向量的加减乘除

Vtor operator + (Vtor A, Vtor B) { return Vtor(A.x + B.x, A.y + B.y); }

Vtor operator - (Point A, Point B) { return Vtor(A.x - B.x, A.y - B.y); }

Vtor operator * (Vtor A, double p) { return Vtor(A.x*p, A.y*p); }

Vtor operator / (Vtor A, double p) { return Vtor(A.x / p, A.y / p); }

bool operator < (Point A, Point B) { return A.x < B.x || (A.x == B.x && A.y < B.y); }

int dcmp(double x){ if (fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }

bool operator == (Point A, Point B) { return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0; }

//向量的点积，长度，夹角

double Dot(Vtor A, Vtor B) { return (A.x*B.x + A.y*B.y); }

double Length(Vtor A) { return sqrt(Dot(A, A)); }

double Angle(Vtor A, Vtor B) { return acos(Dot(A, B) / Length(A) / Length(B)); }

//叉积，三角形面积

double Cross(Vtor A, Vtor B) { return A.x*B.y - A.y*B.x; }

double Area2(Point A, Point B, Point C) { return Cross(B - A, C - A); }

//向量的旋转,求向量的单位法线（即左转90度，然后长度归一）

Vtor Rotate(Vtor A, double rad){ return Vtor(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)); }

Vtor Normal(Vtor A)

{

	double L = Length(A);

	return Vtor(-A.y / L, A.x / L);

}

//直线的交点

Point GetLineIntersection(Point P, Vtor v, Point Q, Vtor w)

{

	Vtor u = P - Q;

	double t = Cross(w, u) / Cross(v, w);

	return P + v*t;

}

//点到直线的距离

double DistanceToLine(Point P, Point A, Point B)

{

	Vtor v1 = B - A;

	return fabs(Cross(P - A, v1)) / Length(v1);

}

//点到线段的距离

double DistanceToSegment(Point P, Point A, Point B)

{

	if (A == B) return Length(P - A);

	Vtor v1 = B - A, v2 = P - A, v3 = P - B;

	if (dcmp(Dot(v1, v2)) < 0) return Length(v2);

	else if (dcmp(Dot(v1, v3)) > 0) return Length(v3);

	else return fabs(Cross(v1, v2)) / Length(v1);

}

//点到直线的映射

Point GetLineProjection(Point P, Point A, Point B)

{

	Vtor v = B - A;

	return A + v*Dot(v, P - A) / Dot(v, v);

}



//判断线段是否规范相交

bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)

{

	double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),

		c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);

	return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;

}

//判断点是否在一条线段上

bool OnSegment(Point P, Point a1, Point a2)

{

	return dcmp(Cross(a1 - P, a2 - P)) == 0 && dcmp(Dot(a1 - P, a2 - P)) < 0;

}

//多边形面积

double PolgonArea(Point *p, int n)

{

	double area = 0;

	for (int i = 1; i < n - 1; ++i)

		area += Cross(p[i] - p[0], p[i + 1] - p[0]);

	return area / 2;

}



struct Line

{

	Point p, b;

	Vtor v;

	Line(){}

	Line(Point a, Point b, Vtor v) : p(a), b(b), v(v) {}

	Line(Point p, Vtor v) : p(p), v(v){}

	Point point(double t) { return p + v*t; }

};

struct Circle

{

	Point c;

	double r;

	Circle(Point tc, double tr) : c(tc), r(tr){}

	Point point(double a)

	{

		return Point(c.x + cos(a)*r, c.y + sin(a)*r);

	}

};

//判断圆与直线是否相交以及求出交点

int getLineCircleIntersection(Line L, Circle C, double &t1, double &t2, vector<Point> &sol)

{

	//    printf(">>>>>>>>>>>>>>>>>>>>>>>>\n");

	//注意sol没有清空哦

	double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;

	double e = a*a + c*c, f = 2 * (a*b + c*d), g = b*b + d*d - C.r*C.r;

	double delta = f*f - 4.0*e*g;

	if (dcmp(delta) < 0) return 0;

	else if (dcmp(delta) == 0)

	{

		t1 = t2 = -f / (2.0*e);

		sol.push_back(L.point(t1));

		return 1;

	}

	t1 = (-f - sqrt(delta)) / (2.0 * e); sol.push_back(L.point(t1));

	t2 = (-f + sqrt(delta)) / (2.0 * e); sol.push_back(L.point(t2));

	return 2;

}

//判断并求出两圆的交点

double angle(Vtor v) { return atan2(v.y, v.x); }

int getCircleIntersection(Circle C1, Circle C2, vector<Point> &sol)

{

	double d = Length(C1.c - C2.c);

	// 圆心重合

	if (dcmp(d) == 0)

	{

		if (dcmp(C1.r - C2.r) == 0) return -1; // 两圆重合

		return 0; // 包含

	}



	// 圆心不重合

	if (dcmp(C1.r + C2.r - d) < 0) return 0; // 相离

	if (dcmp(fabs(C1.r - C2.r) - d) > 0) return 0; // 包含



	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), p2 = C1.point(a + da);

	sol.push_back(p1);

	if (p1 == p2) return 1;

	sol.push_back(p2);

	return 2;

}

//求点到圆的切线

int getTangents(Point p, Circle C, Vtor* v)

{

	Vtor u = C.c - p;

	double dis = Length(u);

	if (dis < C.r)  return 0;

	else if (dcmp(dis - C.r) == 0)

	{

		v[0] = Rotate(u, PI / 2.0);

		return 1;

	}

	else

	{

		double ang = asin(C.r / dis);

		v[0] = Rotate(u, -ang);

		v[1] = Rotate(u, +ang);

		return 2;

	}

}

//求两圆的切线

int getCircleTangents(Circle A, Circle B, Point *a, Point *b)

{

	int cnt = 0;

	if (A.r < B.r) { swap(A, B); swap(a, b); }

	//圆心距的平方

	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);

	double rdiff = A.r - B.r;

	double rsum = A.r + B.r;

	double base = angle(B.c - A.c);

	//重合有无限多条

	if (d2 == 0 && dcmp(A.r - B.r) == 0) return -1;

	//内切

	if (dcmp(d2 - rdiff*rdiff) == 0)

	{

		a[cnt] = A.point(base);

		b[cnt] = B.point(base); cnt++;

		return 1;

	}

	//有外公切线

	double ang = acos((A.r - B.r) / sqrt(d2));

	a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;

	a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;



	//一条内切线

	if (dcmp(d2 - rsum*rsum) == 0)

	{

		a[cnt] = A.point(base); b[cnt] = B.point(PI + base); cnt++;

	}//两条内切线

	else if (dcmp(d2 - rsum*rsum) > 0)

	{

		double ang = acos((A.r + B.r) / sqrt(d2));

		a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;

		a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;

	}

	return cnt;

}





//**********************************

Circle CircumscribedCircle(Point A, Point B, Point C)

{

	Point tmp1 = Point((B.x + C.x) / 2.0, (B.y + C.y) / 2.0);

	Vtor u = C - tmp1;

	u = Rotate(u, PI / 2.0);

	Point tmp2 = Point((A.x + C.x) / 2.0, (A.y + C.y) / 2.0);

	Vtor v = C - tmp2;

	v = Rotate(v, -PI / 2.0);

	Point c = GetLineIntersection(tmp1, u, tmp2, v);

	double r = Length(C - c);

	return Circle(c, r);

}

//得到法向量就得到了这个方向上的向量了

//Circle work1(Point p1, Point p2, Point p3)

// {

//     Vtor nor1 = Normal(p1 - p2);

//     Vtor nor2 = Normal(p2 - p3);

//     Point mid1 = (p1 + p2) / 2.0;

//     Point mid2 = (p2 + p3) / 2.0;

//     Point O = GetLineIntersection(mid1, nor1, mid2, nor2);

//     double r = Length(O - p1);

//     return Circle(O, r);

//}



//不知道为什么我按常规的求法就是不对

//Circle InscribedCircle(Point A,Point B,Point C)

//{

//    Vtor u = A - B;

//    Vtor v = C - B;

//    double ang = Angle(u,v);

//    Vtor vv= Rotate(v,ang / 2.0);

//    u = A - C;

//    v = B - C;

//    ang = Angle(u,v);

//    Vtor uu = Rotate(u,ang / 2.0);

//    Point c = GetLineIntersection(B,vv,C,uu);

//    double r = DistanceToLine(c,A,C);

//    return Circle(c,r);

//}

Circle work2(Point p1, Point p2, Point p3) {

	Vtor v11 = p2 - p1;

	Vtor v12 = p3 - p1;

	Vtor v21 = p1 - p2;

	Vtor v22 = p3 - p2;

	double ang1 = (angle(v11) + angle(v12)) / 2.0;

	double ang2 = (angle(v21) + angle(v22)) / 2.0;

	Vtor vec1 = Vtor(cos(ang1), sin(ang1));

	Vtor vec2 = Vtor(cos(ang2), sin(ang2));

	Point O = GetLineIntersection(p1, vec1, p2, vec2);

	double r = DistanceToLine(O, p1, p2);

	return Circle(O, r);

}

vector<Point> solve4(Point A, Point B, double r, Point C)

{

	Vtor normal = Normal(B - A);

	normal = normal / Length(normal) * r;

	vector<Point> ans;

	double t1 = 0, t2 = 0;

	Vtor tA = A + normal, tB = B + normal;

	getLineCircleIntersection(Line(tA, tB, tB - tA), Circle(C, r), t1, t2, ans);

	tA = A - normal, tB = B - normal;

	getLineCircleIntersection(Line(tA, tB, tB - tA), Circle(C, r), t1, t2, ans);

	return ans;

}

vector<Point> solve5(Point A, Point B, Point C, Point D, double r)

{

	Line lines[5];

	Vtor normal = Normal(B - A) * r;

	Point ta, tb, tc, td;

	ta = A + normal, tb = B + normal;

	lines[0] = Line(ta, tb, tb - ta);

	ta = A - normal, tb = B - normal;

	lines[1] = Line(ta, tb, tb - ta);



	normal = Normal(D - C) * r;

	tc = C + normal, td = D + normal;

	lines[2] = Line(tc, td, td - tc);

	tc = C - normal, td = D - normal;

	lines[3] = Line(tc, td, td - tc);

	vector<Point> ans;

	ans.push_back(GetLineIntersection(lines[0].p, lines[0].v, lines[2].p, lines[2].v));

	ans.push_back(GetLineIntersection(lines[0].p, lines[0].v, lines[3].p, lines[3].v));

	ans.push_back(GetLineIntersection(lines[1].p, lines[1].v, lines[2].p, lines[2].v));

	ans.push_back(GetLineIntersection(lines[1].p, lines[1].v, lines[3].p, lines[3].v));

	return ans;

}

vector<Point> solve6(Circle C1, Circle C2, double r)

{

	vector<Point> vc;

	getCircleIntersection(Circle(C1.c, C1.r + r), Circle(C2.c, C2.r + r), vc);

	return vc;

}



string op;

double x[10];



int main()

{

	//    Read();



	while (cin >> op)

	{

		if (op == "CircumscribedCircle")

		{

			for (int i = 0; i < 6; ++i) cin >> x[i];

			Circle ans = CircumscribedCircle(Point(x[0], x[1]), Point(x[2], x[3]), Point(x[4], x[5]));

			//            Circle ans = work1(Point(x[0],x[1]),Point(x[2],x[3]),Point(x[4],x[5]));

			printf("(%.6lf,%.6lf,%.6lf)\n", ans.c.x, ans.c.y, ans.r);

		}

		else if (op == "InscribedCircle")

		{

			for (int i = 0; i < 6; ++i) cin >> x[i];

			//            Circle ans = InscribedCircle(Point(x[0],x[1]),Point(x[2],x[3]),Point(x[4],x[5]));

			Circle ans = work2(Point(x[0], x[1]), Point(x[2], x[3]), Point(x[4], x[5]));

			printf("(%.6lf,%.6lf,%.6lf)\n", ans.c.x, ans.c.y, ans.r);

		}

		else if (op == "TangentLineThroughPoint")

		{

			for (int i = 0; i < 5; ++i) cin >> x[i];

			Vtor vc[5];

			int len = getTangents(Point(x[3], x[4]), Circle(Point(x[0], x[1]), x[2]), vc);

			double tmp[5];

			for (int i = 0; i < len; ++i)

			{

				double ang = angle(vc[i]);

				if (ang < 0) ang += PI;

				ang = fmod(ang, PI);

				tmp[i] = ang * 180 / PI;

			}

			sort(tmp, tmp + len);

			printf("[");

			for (int i = 0; i < len; ++i)

			{

				printf("%.6lf", tmp[i]);

				if (i != len - 1) printf(",");

			}

			printf("]\n");

		}

		else if (op == "CircleThroughAPointAndTangentToALineWithRadius")

		{

			for (int i = 0; i < 7; ++i) cin >> x[i];

			vector<Point> vc = solve4(Point(x[2], x[3]), Point(x[4], x[5]), x[6], Point(x[0], x[1]));

			sort(vc.begin(), vc.end());

			printf("[");

			for (size_t i = 0; i < vc.size(); ++i)

			{

				printf("(%.6lf,%.6lf)", vc[i].x, vc[i].y);

				if (i != vc.size() - 1) printf(",");

			}

			printf("]\n");

		}

		else if (op == "CircleTangentToTwoLinesWithRadius")

		{

			for (int i = 0; i < 9; ++i) cin >> x[i];

			vector<Point> vc = solve5(Point(x[0], x[1]), Point(x[2], x[3]), Point(x[4], x[5]), Point(x[6], x[7]), x[8]);

			sort(vc.begin(), vc.end());

			printf("[");

			for (size_t i = 0; i < vc.size(); ++i)

			{

				printf("(%.6lf,%.6lf)", vc[i].x, vc[i].y);

				if (i != vc.size() - 1) printf(",");

			}

			printf("]\n");

		}

		else

		{

			for (int i = 0; i < 7; ++i) cin >> x[i];

			vector<Point> vc = solve6(Circle(Point(x[0], x[1]), x[2]), Circle(Point(x[3], x[4]), x[5]), x[6]);

			sort(vc.begin(), vc.end());

			printf("[");

			for (size_t i = 0; i < vc.size(); ++i)

			{

				printf("(%.6lf,%.6lf)", vc[i].x, vc[i].y);

				if (i != vc.size() - 1) printf(",");

			}

			printf("]\n");

		}

	}

}