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"); } } }