• 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 fabs(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 }
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  • 原文地址:https://www.cnblogs.com/St-Lovaer/p/11439816.html
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