http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3726
题意:
新白书p267, 有说
给出三角形三点求外接圆,内接圆,给出一点求以及圆求过该点的切线, 给出一直线和一个点求过该点与直线想切的圆,圆半径给出。给出两条相交的直线求与这两条直线想切的圆, 给出两个相离的圆,求与这两个圆都想切的圆
思路:
其实没什么很难的就是模板运用,还要注意细节什么的,这题考了很多二维几何的模板,值得一做,话说这是做ACM 题目以来写的最长的题目。细心。。
#include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <string> #include <set> #include <functional> #include <numeric> #include <sstream> #include <stack> #include <map> #include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define PI acos(-1.0) #define L(x) (x) << 1 #define R(x) (x) << 1 | 1 #define MID(l, r) (l + r) >> 1 #define Min(x, y) (x) < (y) ? (x) : (y) #define Max(x, y) (x) < (y) ? (y) : (x) #define E(x) (1 << (x)) #define iabs(x) (x) < 0 ? -(x) : (x) #define OUT(x) printf("%I64d ", x) #define lowbit(x) (x)&(-x) #define Read() freopen("data.in", "r", stdin) #define Write() freopen("d.out", "w", stdout) #define ll unsigned long long #define M 100007 #define N 65736 using namespace std; 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(">>>>>>>>>>>>>>>>>>>>>>>> "); //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) ",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) ",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("] "); } 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("] "); } 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("] "); } 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("] "); } } return 0; }