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 65736const 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");}}}