计算几何

  1 const  double eps=1e-10;
  2 const double PI=acos(-1.0);
  3 using namespace std;
  4 struct Point{
  5     double x;
  6     double y;
  7     Point(double x=0,double y=0):x(x),y(y){}
  8     void operator<<(Point &A) {cout<<A.x<<‘ ‘<<A.y<<endl;}
  9 };
 10  
 11 int dcmp(double x)  {return (x>eps)-(x<-eps); }
 12 int sgn(double x)  {return (x>eps)-(x<-eps); }
 13 typedef  Point  Vector;
 14  
 15 Vector  operator +(Vector A,Vector B) { return Vector(A.x+B.x,A.y+B.y);}
 16 Vector  operator -(Vector A,Vector B) { return Vector(A.x-B.x,A.y-B.y); }
 17 Vector  operator *(Vector A,double p) { return Vector(A.x*p,A.y*p);  }
 18 Vector  operator /(Vector A,double p) {return Vector(A.x/p,A.y/p);}
 19 ostream &operator<<(ostream & out,Point & P) { out<<P.x<<‘ ‘<<P.y<<endl; return out;}
 20 bool  operator< (const Point &A,const Point &B) { return dcmp(A.x-B.x)<0||(dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)<0); }
 21 bool  operator== ( const Point &A,const Point &B) { return dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)==0;}
 22  
 23 double  Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}
 24 double  Cross(Vector A,Vector B)  {return A.x*B.y-B.x*A.y; }
 25 double  Length(Vector A)  { return sqrt(Dot(A, A));}
 26 double  Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}
 27 double  Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}
 28 
 29 Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}
 30 Vector Normal(Vector A) {double L=Length(A);return Vector(-A.y/L,A.x/L);}
 31  
 32 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){
 33     Vector u=P-Q;
 34     double t=Cross(w, u)/Cross(v,w);
 35     return P+v*t;
 36 }//输入两个点斜式方程输出交点 
 37  
 38 double DistanceToLine(Point P,Point A,Point B){
 39     Vector v1=P-A; Vector v2=B-A;
 40     return fabs(Cross(v1,v2))/Length(v2);
 41 }//输入三个点，输出P到直线AB的距离 
 42  
 43 double DistanceToSegment(Point P,Point A,Point B){
 44     if(A==B)  return Length(P-A);
 45     Vector v1=B-A;
 46     Vector v2=P-A;
 47     Vector v3=P-B;
 48     if(dcmp(Dot(v1,v2))==-1)    return  Length(v2);
 49     else if(Dot(v1,v3)>0)    return Length(v3);
 50     else return DistanceToLine(P, A, B);
 51 }//输入三个点，输出P到线段AB的距离 
 52  
 53 Point GetLineProjection(Point P,Point A,Point B){
 54     Vector v=B-A;
 55     Vector v1=P-A;
 56     double t=Dot(v,v1)/Dot(v,v);
 57     return  A+v*t;
 58 }//输入三个点，输出P在直线AB上的投影点 
 59  
 60 bool  SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){
 61     double c1=Cross(b1-a1, a2-a1);
 62     double c2=Cross(b2-a1, a2-a1);
 63     double c3=Cross(a1-b1, b2-b1);
 64     double c4=Cross(a2-b1, b2-b1);
 65     return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
 66 }//输入四个点，判断两条线段是否有交点 
 67  
 68 bool  OnSegment(Point P,Point A,Point B){
 69     return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0;
 70 }//输入三个点，判断P是否在线段AB上 
 71  
 72 double PolygonArea(Point *p,int n){
 73     double area=0;
 74     for(int i=1;i<n-1;i++){
 75         area+=Cross(p[i]-p[0], p[i+1]-p[0]); 
 76     }
 77     return area/2;
 78 }//输入p[i]首地址和p[]的数量，（p[0]到p[n-1]），输出多边形面
 79  
 80 Point read_point(){
 81     Point P;
 82     scanf("%lf%lf",&P.x,&P.y);
 83     return P;
 84 }//输入一个点 
 85  
 86 // ---------------与圆有关的--------
 87  
 88 struct Circle{
 89     Point c;
 90     double r;
 91     Circle(Point c=Point(0,0),double r=0):c(c),r(r) {}
 92     Point point(double a){
 93         return Point(c.x+r*cos(a),c.y+r*sin(a));
 94     }  
 95 };
 96  
 97 struct Line{
 98     Point p;
 99     Vector v;
100     Line(Point p=Point(0,0),Vector v=Vector(0,1)):p(p),v(v) {}
101     Point point(double t){
102         return Point(p+v*t);
103     }
104 };
105  
106 int getLineCircleIntersection(Line L,Circle C,double &t1,double &t2,vector<Point> &sol){
107     double a=L.v.x;
108     double b=L.p.x-C.c.x;
109     double c=L.v.y;
110     double d=L.p.y-C.c.y;
111     double e=a*a+c*c;
112     double f=2*(a*b+c*d);
113     double g=b*b+d*d-C.r*C.r;
114     double delta=f*f-4*e*g;
115     
116     if(dcmp(delta)<0) return 0;
117     if(dcmp(delta)==0){
118         t1=t2=-f/(2*e);
119         sol.push_back(L.point(t1));
120         return 1;
121     }
122     
123     else{
124         t1=(-f-sqrt(delta))/(2*e);
125         t2=(-f+sqrt(delta))/(2*e);
126         
127         sol.push_back(L.point(t1));
128         sol.push_back(L.point(t2));
129         
130         return 2;
131     } 
132 }
133  
134 // 向量极角公式
135  
136 double angle(Vector v)  {return atan2(v.y,v.x);}
137  
138 int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point> &sol)
139 {
140     double d=Length(C1.c-C2.c);
141     
142     if(dcmp(d)==0)
143     {
144         if(dcmp(C1.r-C2.r)==0)  return -1;  // 重合
145         else return 0;    //  内含  0 个公共点
146     }
147     
148     if(dcmp(C1.r+C2.r-d)<0)  return 0;  // 外离
149     if(dcmp(fabs(C1.r-C2.r)-d)>0)  return 0;  // 内含
150     
151     double a=angle(C2.c-C1.c);
152     double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d));
153     
154     Point p1=C1.point(a-da);
155     Point p2=C1.point(a+da);
156     
157     sol.push_back(p1);
158     
159     if(p1==p2)  return 1; // 相切
160     else{
161         sol.push_back(p2);
162         return 2;
163     }
164 }
165  
166  
167 //  求点到圆的切线
168  
169 int getTangents(Point p,Circle C,Vector *v){
170     Vector u=C.c-p;
171     
172     double dist=Length(u);
173     
174     if(dcmp(dist-C.r)<0)  return 0;
175     
176     else if(dcmp(dist-C.r)==0){
177         v[0]=Rotate(u,PI/2);
178         return 1;
179     }
180     
181     else{
182         double ang=asin(C.r/dist);
183         v[0]=Rotate(u,-ang);
184         v[1]=Rotate(u,+ang);
185         return 2;
186     }
187 }
188  
189 //  求两圆公切线
190 int getTangents(Circle A,Circle B,Point *a,Point *b){
191     int cnt=0;
192     
193     if(A.r<B.r){
194         swap(A,B); swap(a, b);  //  有时需标记
195     }
196     
197     double d=Length(A.c-B.c);
198     double rdiff=A.r-B.r;
199     double rsum=A.r+B.r;
200     
201     if(dcmp(d-rdiff)<0)  return 0;   // 内含
202     
203     double base=angle(B.c-A.c);
204     if(dcmp(d)==0&&dcmp(rdiff)==0)   return -1 ;  // 重合 无穷多条切线
205     
206     if(dcmp(d-rdiff)==0)             // 内切   外公切线
207     {
208         a[cnt]=A.point(base);
209         b[cnt]=B.point(base);
210         cnt++;
211         return 1;
212     }
213     
214     // 有外公切线的情形
215     
216     double ang=acos(rdiff/d);
217     a[cnt]=A.point(base+ang);
218     b[cnt]=B.point(base+ang);
219     cnt++;
220     a[cnt]=A.point(base-ang);
221     b[cnt]=B.point(base-ang);
222     cnt++;
223     
224     if(dcmp(d-rsum)==0)     // 外切 有内公切线
225     {
226         a[cnt]=A.point(base);
227         b[cnt]=B.point(base+PI);
228         cnt++;
229     }
230     
231     else  if(dcmp(d-rsum)>0)   // 外离   又有两条外公切线
232     {
233         double  ang_in=acos(rsum/d);
234         a[cnt]=A.point(base+ang_in);
235         b[cnt]=B.point(base+ang_in+PI);
236         cnt++;
237         a[cnt]=A.point(base-ang_in);
238         b[cnt]=B.point(base-ang_in+PI);
239         cnt++;
240     }
241     
242     return cnt;
243 }
244  
245 Point Zero=Point(0,0);
246 
247 double TriAngleCircleInsection(Circle C, Point A, Point B){
248     Vector OA = A-C.c, OB = B-C.c;
249     Vector BA = A-B, BC = C.c-B;
250     Vector AB = B-A, AC = C.c-A;
251     double DOA = Length(OA), DOB = Length(OB),DAB = Length(AB), r = C.r;
252     if(dcmp(Cross(OA,OB)) == 0) return 0;
253     if(dcmp(DOA-C.r) < 0 && dcmp(DOB-C.r) < 0) return Cross(OA,OB)*0.5;
254     else if(DOB < r && DOA >= r) {
255         double x = (Dot(BA,BC) + sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
256         double TS = Cross(OA,OB)*0.5;
257         return asin(TS*(1-x/DAB)*2/r/DOA)*r*r*0.5+TS*x/DAB;
258     }
259     else if(DOB >= r && DOA < r) {
260         double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
261         double TS = Cross(OA,OB)*0.5;
262         return asin(TS*(1-y/DAB)*2/r/DOB)*r*r*0.5+TS*y/DAB;
263     }
264     else if(fabs(Cross(OA,OB)) >= r*DAB || Dot(AB,AC) <= 0 || Dot(BA,BC) <= 0) {
265         if(Dot(OA,OB) < 0){
266             if(Cross(OA,OB) < 0) return (-acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
267             else return ( acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
268         }
269         else return asin(Cross(OA,OB)/DOA/DOB)*r*r*0.5;
270     }
271     else {
272         double x = (Dot(BA,BC)+sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
273         double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
274         double TS = Cross(OA,OB)*0.5;
275         return
276 (asin(TS*(1-x/DAB)*2/r/DOA)+asin(TS*(1-y/DAB)*2/r/DOB))*r*r*0.5 + TS*((x+y)/DAB-1);
277     }
278 }