poj2986&&poj3675【圆与多边形相交面积模板】

poj3675求圆心在原点的园与多边形相交的面积表示不懂以后慢慢理解吧；

#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>#include<vector>const  double eps=1e-10;const double PI=acos(-1.0);using namespace std;struct Point{    double x;    double y;    Point(double x=0,double y=0):x(x),y(y){}    void operator<<(Point &A) {cout<<A.x<<' '<<A.y<<endl;}};int dcmp(double x)  {return (x>eps)-(x<-eps); }int sgn(double x)  {return (x>eps)-(x<-eps); }typedef  Point  Vector;Vector  operator +(Vector A,Vector B) { return Vector(A.x+B.x,A.y+B.y);}Vector  operator -(Vector A,Vector B) { return Vector(A.x-B.x,A.y-B.y); }Vector  operator *(Vector A,double p) { return Vector(A.x*p,A.y*p);  }Vector  operator /(Vector A,double p) {return Vector(A.x/p,A.y/p);}ostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y<<endl; return out;}//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); }bool  operator== ( const Point &A,const Point &B) { return dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)==0;}double  Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}double  Cross(Vector A,Vector B)  {return A.x*B.y-B.x*A.y; }double  Length(Vector A)  { return sqrt(Dot(A, A));}double  Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}double  Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A) {double L=Length(A);return Vector(-A.y/L,A.x/L);}Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){    Vector u=P-Q;    double t=Cross(w, u)/Cross(v,w);    return P+v*t;    }double DistanceToLine(Point P,Point A,Point B){    Vector v1=P-A; Vector v2=B-A;    return fabs(Cross(v1,v2))/Length(v2);    }double DistanceToSegment(Point P,Point A,Point B){    if(A==B)  return Length(P-A);        Vector v1=B-A;    Vector v2=P-A;    Vector v3=P-B;        if(dcmp(Dot(v1,v2))==-1)    return  Length(v2);    else if(Dot(v1,v3)>0)    return Length(v3);        else return DistanceToLine(P, A, B);    }Point GetLineProjection(Point P,Point A,Point B){    Vector v=B-A;    Vector v1=P-A;    double t=Dot(v,v1)/Dot(v,v);        return  A+v*t;}bool  SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){    double c1=Cross(b1-a1, a2-a1);    double c2=Cross(b2-a1, a2-a1);    double c3=Cross(a1-b1, b2-b1);    double c4=Cross(a2-b1, b2-b1);        return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0 ;    }bool  OnSegment(Point P,Point A,Point B){    return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0;}double PolygonArea(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;    }Point  read_point(){    Point P;    scanf("%Lf%Lf",&P.x,&P.y);    return  P;}// ---------------与圆有关的--------struct Circle{    Point c;    double r;        Circle(Point c=Point(0,0),double r=0):c(c),r(r) {}        Point point(double a)    {        return Point(c.x+r*cos(a),c.y+r*sin(a));    }        };struct  Line{    Point p;    Vector v;    Line(Point p=Point(0,0),Vector v=Vector(0,1)):p(p),v(v) {}        Point point(double t)    {        return Point(p+v*t);    }    };int getLineCircleIntersection(Line L,Circle C,double &t1,double &t2,vector<Point> &sol){    double a=L.v.x;    double b=L.p.x-C.c.x;    double c=L.v.y;    double d=L.p.y-C.c.y;        double e=a*a+c*c;    double f=2*(a*b+c*d);    double g=b*b+d*d-C.r*C.r;        double delta=f*f-4*e*g;        if(dcmp(delta)<0) return 0;        if(dcmp(delta)==0)    {        t1=t2=-f/(2*e);        sol.push_back(L.point(t1));        return 1;    }        else    {        t1=(-f-sqrt(delta))/(2*e);        t2=(-f+sqrt(delta))/(2*e);                sol.push_back(L.point(t1));        sol.push_back(L.point(t2));                return 2;    }    }// 向量极角公式double angle(Vector v)  {return atan2(v.y,v.x);}int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point> &sol){    double d=Length(C1.c-C2.c);        if(dcmp(d)==0)    {        if(dcmp(C1.r-C2.r)==0)  return -1;  // 重合        else return 0;    //  内含  0 个公共点    }        if(dcmp(C1.r+C2.r-d)<0)  return 0;  // 外离    if(dcmp(fabs(C1.r-C2.r)-d)>0)  return 0;  // 内含        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);    Point p2=C1.point(a+da);        sol.push_back(p1);        if(p1==p2)  return 1; // 相切    else    {        sol.push_back(p2);        return 2;    }}//  求点到圆的切线int getTangents(Point p,Circle C,Vector *v){    Vector u=C.c-p;        double dist=Length(u);        if(dcmp(dist-C.r)<0)  return 0;        else if(dcmp(dist-C.r)==0)    {        v[0]=Rotate(u,PI/2);        return 1;    }        else    {                double ang=asin(C.r/dist);        v[0]=Rotate(u,-ang);        v[1]=Rotate(u,+ang);        return 2;    }    }//  求两圆公切线int getTangents(Circle A,Circle B,Point *a,Point *b){    int cnt=0;        if(A.r<B.r)    {        swap(A,B); swap(a, b);  //  有时需标记    }        double d=Length(A.c-B.c);        double rdiff=A.r-B.r;    double rsum=A.r+B.r;        if(dcmp(d-rdiff)<0)  return 0;   // 内含        double base=angle(B.c-A.c);        if(dcmp(d)==0&&dcmp(rdiff)==0)   return -1 ;  // 重合 无穷多条切线        if(dcmp(d-rdiff)==0)             // 内切   外公切线    {        a[cnt]=A.point(base);        b[cnt]=B.point(base);        cnt++;        return 1;    }        // 有外公切线的情形        double ang=acos(rdiff/d);    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++;        if(dcmp(d-rsum)==0)     // 外切 有内公切线    {        a[cnt]=A.point(base);        b[cnt]=B.point(base+PI);        cnt++;    }        else  if(dcmp(d-rsum)>0)   // 外离   又有两条外公切线    {        double  ang_in=acos(rsum/d);        a[cnt]=A.point(base+ang_in);        b[cnt]=B.point(base+ang_in+PI);        cnt++;        a[cnt]=A.point(base-ang_in);        b[cnt]=B.point(base-ang_in+PI);        cnt++;    }        return cnt;}Point Zero=Point(0,0);double  common_area(Circle C,Point A,Point B){   // if(A==B)  return 0;    if(A==C.c||B==C.c)  return 0;    double  OA=Length(A-C.c),OB=Length(B-C.c);    double  d=DistanceToLine(Zero, A, B);    int sg=sgn(Cross(A,B));    if(sg==0)  return 0;    double angle=Angle(A,B);    if(dcmp(OA-C.r)<=0&&dcmp(OB-C.r)<=0)    {        return Cross(A,B)/2;    }    else if(dcmp(OA-C.r)>=0&&dcmp(OB-C.r)>=0&&dcmp(d-C.r)>=0)    {        return  sg*C.r*C.r*angle/2;    }    else if (dcmp(OA-C.r)>=0&&dcmp(OB-C.r)>=0&&dcmp(d-C.r)<0)    {        Point prj=GetLineProjection(Zero, A, B);        if(OnSegment(prj, A, B))        {        vector<Point> p;        Line L=Line(A,B-A);        double t1,t2;        getLineCircleIntersection(L,C, t1, t2, p);        double s1=0;        s1=C.r*C.r*angle/2;        double s2=0;        s2=C.r*C.r*Angle(p[0],p[1])/2;        s2-=fabs(Cross(p[0],p[1])/2);        s1=s1-s2;        return  sg*s1;        }        else        {            return sg*C.r*C.r*angle/2;        }    }    else    {                   if(dcmp(OB-C.r)<0)            {                               Point temp=A;                A=B;                B=temp;            }         Point inter_point;                    double t1,t2;            Line L=Line(A,B-A);            vector<Point> inter;            getLineCircleIntersection(L, C, t1, t2,inter);                            if(OnSegment(inter[0], A, B))                inter_point=inter[0];            else            {                inter_point=inter[1];            }        //        两种方法求交点都可以//        Point  prj=GetLineProjection(Zero, A, B);//        //        Vector v=B-A;//        v=v/Length(v);//        double mov=sqrt(C.r*C.r-d*d);//        //        if(OnSegment(prj+v*mov, A, B))//        {//            inter_point=prj+v*mov;//        }//        else//        {//            inter_point=prj+Vector(-v.x,-v.y)*mov;//        }                            double s=fabs(Cross(inter_point, A)/2);            s+=C.r*C.r*Angle(inter_point,B)/2;            return s*sg;    }}int main(){    Point p[60];    double R;    int n;    Circle C(Zero,0);    while(cin>>R)    {        C.r=R;        cin>>n;        for(int i=0;i<n;i++)            p[i]=read_point();                p[n]=p[0];        double ans=0;                for(int i=0;i<n;i++)            ans+=common_area(C, p[i], p[i+1]);        printf("%.2f\n",fabs(ans)+eps);    }         return 0;}

比较好理解：

//多边形与圆点相交面积 poj3675#include <iostream>#include <stdio.h>#include <cmath>using namespace std;struct point{    double x,y;}a[55];double r;//半径double dist_1point(double x0,double y0)//点到原点距离{    return sqrt(x0*x0+y0*y0);}double dist_2point(double x1,double y1,double x2,double y2) //两点距离{    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}double dist_line(double x1,double y1,double x2,double y2)//直线到原点距离{    double A,B,C,dist;    A=y1-y2;    B=x1-x2;    C=x1*y2-x2*y1;    dist=fabs(C)/sqrt(A*A+B*B);//直线到原点距离公式    return dist;}double get_cos(double a,double b,double c)  //余弦定理求角度{double angel=(b*b+c*c-a*a)/(2*b*c);    return angel;}point get_point(double x0,double y0)//计算点与原点的直线与圆相交的交点{    double k;    point temp;    if(x0!=0)//若斜率存在    {        k=y0/x0;        temp.x=fabs(r)/sqrt(1+k*k);//判断是两个点的哪一个if(x0<0) temp.x=-temp.x;//temp.x应该与x0同符号        temp.y=k*temp.x;    }    else//斜率不存在    {        temp.x=0;if(y0>0) temp.y=r;//判断是两个点的哪一个else temp.y=-r;//temp.y应该与y0同符号    }    return temp;}int fi(double x1,double y1,double x2,double y2)//关键啊啊啊！！！！{    if (x1*y2-x2*y1>0) return 1;//判断是相加还是相减    else return -1;}double get_area(double x1,double y1,double x2,double y2) //三角剖分{int sign=fi(x1,y1,x2,y2);//判断三角形面积加还是减    double s;//总面积    double l=dist_line(x1,y1,x2,y2);   //l = 直线ab与原点距离    double a=dist_1point(x1,y1);       //a = 线段a长度    double b=dist_1point(x2,y2);   //b = 线段b长度    double c=dist_2point(x1,y1,x2,y2);     //c = 线段c长度if(a==0 || b==0)return 0;//若其中一条边为0，返回面积0//第一种情况：三角形的两条边a、b全部短于半径。    if(a<=r && b<=r)    {        s=fabs(x1*y2-x2*y1)/2.0;        return s*sign;    }    //第二种情况：a、b两条边长于半径，l也长于半径。    else if(a>=r && b>=r && l>=r)    {        point t1=get_point(x1,y1);        point t2=get_point(x2,y2);        double d=dist_2point(t1.x,t1.y,t2.x,t2.y);        double sita1=acos(get_cos(d,r,r));        double s=fabs(sita1*r*r/2.0); //扇形面积：s=θ*r*r/2        return s*sign;    }//第三种情况：a、b两条边长于半径，但l短于半径，并且垂足落在这条边上。    else if(a>=r && b>=r && l<=r && (get_cos(a,b,c)<=0 || get_cos(b,a,c)<=0))    {        point t1=get_point(x1,y1);        point t2=get_point(x2,y2);        double d=dist_2point(t1.x,t1.y,t2.x,t2.y);        double sita=acos(get_cos(d,r,r));        s=fabs(sita*r*r/2.0);        return s*sign;    }//第四种情况：a、b两条边长于半径，但l短于半径，且垂足没有落在这条边上。    else if(a>=r && b>=r && l<=r && (get_cos(a,b,c)>0 && get_cos(b,a,c)>0))    {double xx1,xx2,yy1,yy2;//点(x1,y1)与(x2,y2)组成的直线与圆的交点if(x1!=x2)//若斜率存在{double k12=(y1-y2)/(x1-x2);double b12=y1-k12*x1;double a0=(1+k12*k12);double b0=(2*k12*b12);double c0=(b12*b12-r*r);//化成一元二次方程，用公式求出两个交点xx1=(-b0+sqrt(b0*b0-4*a0*c0))/(2*a0);yy1=k12*xx1+b12;xx2=(-b0-sqrt(b0*b0-4*a0*c0))/(2*a0);yy2=k12*xx2+b12;}else//若斜率不存在 x1==x2{xx1=x1;xx2=x1;yy1=sqrt(r*r-x1*x1);yy2=-sqrt(r*r-x1*x1);}point t1=get_point(x1,y1);//(x1,y1),(0,0)组成直线与圆的交点point t2=get_point(x2,y2);//(x2,y2),(0,0)组成直线与圆的交点double d1=dist_2point(xx1,yy1,xx2,yy2);//直线1与原点距离double d2=dist_2point(t1.x,t1.y,t2.x,t2.y);//直线2与原点距离double sita1=acos(get_cos(d1,r,r));  //小的扇形弧度double sita2=acos(get_cos(d2,r,r)); //大的扇形弧度double s1=fabs(sita1*r*r/2.0);   //小的扇形面积double s2=fabs(sita2*r*r/2.0);   //大的扇形面积double s3=fabs(xx1*yy2-xx2*yy1)/2.0;  //三角形面积        s=s2+s3-s1;//相交面积        return s*sign;    }//第五种情况1：三角形的两条边一条长于半径，另外一条短于半径    else if(a>=r && b<=r)//a长于半径，b短于半径    {double xxx,yyy;if(x1!=x2)//斜率存在        {double k12=(y1-y2)/(x1-x2);double b12=y1-k12*x1;double a0=(1+k12*k12);double b0=(2*k12*b12);double c0=(b12*b12-r*r);//化成一元二次方程，用公式求出两个交点double xx1=(-b0+sqrt(b0*b0-4*a0*c0))/(2*a0);double yy1=k12*xx1+b12;double xx2=(-b0-sqrt(b0*b0-4*a0*c0))/(2*a0);double yy2=k12*xx2+b12;//判断两个交点中的哪一个，应在(x1,x2)两点之间if(x1<=xx1 && xx1<=x2 || x2<=xx1 && xx1<=x1) {xxx=xx1; yyy=yy1;}else {xxx=xx2; yyy=yy2;}}else//斜率不存在 x1==x2{double xx1=x1;double yy1=-sqrt(r*r-x1*x1);double yy2=sqrt(r*r-x1*x1);//判断两个交点中的哪一个，应在(y1,y2)两点之间if(y1<=yy1 && yy1<=y2 || y2<=yy1 && yy1<=y1) {yyy=yy1; xxx=xx1;}else {yyy=yy2; xxx=xx1;}}        //判断交点(该点已判断方向)        point t1=get_point(x1,y1);        double ddd=dist_2point(t1.x,t1.y,xxx,yyy);        double sita1=acos(get_cos(ddd,r,r));        double s1=fabs(sita1*r*r/2.0);        double s3=fabs(xxx*y2-yyy*x2)/2.0;        s=s1+s3;//相交面积        return s*sign;    }//第五种情况2：三角形的两条边一条长于半径，另外一条短于半径，与上述同理！！！    else if(a<=r && b>=r)//a短于半径，b长于半径    {double xxx,yyy;//与上述同理！！！if(x1-x2!=0){double k12=(y1-y2)/(x1-x2);double b12=y1-k12*x1;double a0=(1+k12*k12);double b0=(2*k12*b12);double c0=(b12*b12-r*r);double xx1=(-b0+sqrt(b0*b0-4*a0*c0))/(2*a0);double yy1=k12*xx1+b12;double xx2=(-b0-sqrt(b0*b0-4*a0*c0))/(2*a0);double yy2=k12*xx2+b12;if(x1<=xx1 && xx1<=x2 || x2<=xx1 && xx1<=x1) {xxx=xx1; yyy=yy1;}else {xxx=xx2; yyy=yy2;}}else{double yy1=-sqrt(r*r-x1*x1);double yy2=sqrt(r*r-x1*x1);double xx1=x1;if(y1<=yy1 && yy1<=y2 || y2<=yy1 && yy1<=y1) {yyy=yy1; xxx=xx1;}else {yyy=yy2; xxx=xx1;}}        point t1=get_point(x2,y2);        double ddd=dist_2point(t1.x,t1.y,xxx,yyy);        double sita1=acos(get_cos(ddd,r,r));        double s1=fabs(sita1*r*r/2.0);        double s3=fabs(xxx*y1-yyy*x1)/2.0;        s=s1+s3;        return s*sign;    }else return 0;}int main(){    int i,n;double area;    while(scanf("%lf",&r)!=EOF)    {        scanf("%d",&n);        for(i=0;i<n;i++)        {            scanf("%lf%lf",&a[i].x,&a[i].y);        }        a[n]=a[0]; area=0;        for(i=0;i<n;i++)//原点与其中两个点组成三角形来判断        {            area+=get_area(a[i].x,a[i].y,a[i+1].x,a[i+1].y);        }        printf("%.2lf\n",fabs(area));    }    return 0;}

poj2986求圆与三角形相交的面积

#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>#include<vector>const  double eps=1e-10;const long double PI=acos(-1.0);using namespace std;struct Point{    long double x;    long double y;    Point(long double x=0,long double y=0):x(x),y(y){}    void operator<<(Point &A) {cout<<A.x<<' '<<A.y<<endl;}};int dcmp(long double x)  {return (x>eps)-(x<-eps); }int sgn(long double x)  {return (x>eps)-(x<-eps); }typedef  Point  Vector;Vector  operator +(Vector A,Vector B) { return Vector(A.x+B.x,A.y+B.y);}Vector  operator -(Vector A,Vector B) { return Vector(A.x-B.x,A.y-B.y); }Vector  operator *(Vector A,long double p) { return Vector(A.x*p,A.y*p);  }Vector  operator /(Vector A,long double p) {return Vector(A.x/p,A.y/p);}ostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y<<endl; return out;}//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); }bool  operator== ( const Point &A,const Point &B) { return dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)==0;}long double  Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}long double  Cross(Vector A,Vector B)  {return A.x*B.y-B.x*A.y; }long double  Length(Vector A)  { return sqrt(Dot(A, A));}long double  Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}long double  Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}Vector Rotate(Vector A,long double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A) {long double L=Length(A);return Vector(-A.y/L,A.x/L);}Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){    Vector u=P-Q;    long double t=Cross(w, u)/Cross(v,w);    return P+v*t;    }long double DistanceToLine(Point P,Point A,Point B){    Vector v1=P-A; Vector v2=B-A;    return fabs(Cross(v1,v2))/Length(v2);    }long double DistanceToSegment(Point P,Point A,Point B){    if(A==B)  return Length(P-A);        Vector v1=B-A;    Vector v2=P-A;    Vector v3=P-B;        if(dcmp(Dot(v1,v2))==-1)    return  Length(v2);    else if(Dot(v1,v3)>0)    return Length(v3);        else return DistanceToLine(P, A, B);    }Point GetLineProjection(Point P,Point A,Point B){    Vector v=B-A;    Vector v1=P-A;    long double t=Dot(v,v1)/Dot(v,v);        return  A+v*t;}bool  SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){    long double c1=Cross(b1-a1, a2-a1);    long double c2=Cross(b2-a1, a2-a1);    long double c3=Cross(a1-b1, b2-b1);    long double c4=Cross(a2-b1, b2-b1);        return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0 ;    }bool  OnSegment(Point P,Point A,Point B){    return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0;}long double PolygonArea(Point *p,int n){    long 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;    }Point  read_point(){    Point P;    scanf("%Lf%Lf",&P.x,&P.y);    return  P;}// ---------------与圆有关的--------struct Circle{    Point c;    long double r;        Circle(Point c=Point(0,0),long double r=0):c(c),r(r) {}        Point point(long double a)    {        return Point(c.x+r*cos(a),c.y+r*sin(a));    }        };struct  Line{    Point p;    Vector v;    Line(Point p=Point(0,0),Vector v=Vector(0,1)):p(p),v(v) {}        Point point(long double t)    {        return Point(p+v*t);    }    };int getLineCircleIntersection(Line L,Circle C,long double &t1,long double &t2,vector<Point> &sol){    long double a=L.v.x;    long double b=L.p.x-C.c.x;    long double c=L.v.y;    long double d=L.p.y-C.c.y;        long double e=a*a+c*c;    long double f=2*(a*b+c*d);    long double g=b*b+d*d-C.r*C.r;        long double delta=f*f-4*e*g;        if(dcmp(delta)<0) return 0;        if(dcmp(delta)==0)    {        t1=t2=-f/(2*e);        sol.push_back(L.point(t1));        return 1;    }        else    {        t1=(-f-sqrt(delta))/(2*e);        t2=(-f+sqrt(delta))/(2*e);                sol.push_back(L.point(t1));        sol.push_back(L.point(t2));                return 2;    }    }// 向量极角公式long double angle(Vector v)  {return atan2(v.y,v.x);}int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point> &sol){    long double d=Length(C1.c-C2.c);        if(dcmp(d)==0)    {        if(dcmp(C1.r-C2.r)==0)  return -1;  // 重合        else return 0;    //  内含  0 个公共点    }        if(dcmp(C1.r+C2.r-d)<0)  return 0;  // 外离    if(dcmp(fabs(C1.r-C2.r)-d)>0)  return 0;  // 内含        long double a=angle(C2.c-C1.c);    long double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d));        Point p1=C1.point(a-da);    Point p2=C1.point(a+da);        sol.push_back(p1);        if(p1==p2)  return 1; // 相切    else    {        sol.push_back(p2);        return 2;    }}//  求点到圆的切线int getTangents(Point p,Circle C,Vector *v){    Vector u=C.c-p;        long double dist=Length(u);        if(dcmp(dist-C.r)<0)  return 0;        else if(dcmp(dist-C.r)==0)    {        v[0]=Rotate(u,PI/2);        return 1;    }        else    {                long double ang=asin(C.r/dist);        v[0]=Rotate(u,-ang);        v[1]=Rotate(u,+ang);        return 2;    }    }//  求两圆公切线int getTangents(Circle A,Circle B,Point *a,Point *b){    int cnt=0;        if(A.r<B.r)    {        swap(A,B); swap(a, b);  //  有时需标记    }        long double d=Length(A.c-B.c);        long double rdiff=A.r-B.r;    long double rsum=A.r+B.r;        if(dcmp(d-rdiff)<0)  return 0;   // 内含        long double base=angle(B.c-A.c);        if(dcmp(d)==0&&dcmp(rdiff)==0)   return -1 ;  // 重合 无穷多条切线        if(dcmp(d-rdiff)==0)             // 内切   外公切线    {        a[cnt]=A.point(base);        b[cnt]=B.point(base);        cnt++;        return 1;    }        // 有外公切线的情形        long double ang=acos(rdiff/d);    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++;        if(dcmp(d-rsum)==0)     // 外切 有内公切线    {        a[cnt]=A.point(base);        b[cnt]=B.point(base+PI);        cnt++;    }        else  if(dcmp(d-rsum)>0)   // 外离   又有两条外公切线    {        long double  ang_in=acos(rsum/d);        a[cnt]=A.point(base+ang_in);        b[cnt]=B.point(base+ang_in+PI);        cnt++;        a[cnt]=A.point(base-ang_in);        b[cnt]=B.point(base-ang_in+PI);        cnt++;    }        return cnt;}int n;Point Zero=Point(0,0);long double  common_area(Circle C,Point A,Point B){   // if(A==B)  return 0;    if(A==C.c||B==C.c)  return 0;    long double  OA=Length(A-C.c),OB=Length(B-C.c);    long double  d=DistanceToLine(Zero, A, B);    int sg=sgn(Cross(A,B));    if(sg==0)  return 0;    long double angle=Angle(A,B);    if(dcmp(OA-C.r)<=0&&dcmp(OB-C.r)<=0)    {        return Cross(A,B)/2;    }    else if(dcmp(OA-C.r)>=0&&dcmp(OB-C.r)>=0&&dcmp(d-C.r)>=0)    {        return  sg*C.r*C.r*angle/2;    }    else if (dcmp(OA-C.r)>=0&&dcmp(OB-C.r)>=0&&dcmp(d-C.r)<0)    {        Point prj=GetLineProjection(Zero, A, B);        if(OnSegment(prj, A, B))        {        vector<Point> p;        Line L=Line(A,B-A);        long double t1,t2;        getLineCircleIntersection(L,C, t1, t2, p);        long double s1=0;        s1=C.r*C.r*angle/2;        long double s2=0;        s2=C.r*C.r*Angle(p[0],p[1])/2;        s2-=fabs(Cross(p[0],p[1])/2);        s1=s1-s2;        return  sg*s1;        }        else        {            return sg*C.r*C.r*angle/2;        }    }    else    {                   if(dcmp(OB-C.r)<0)            {                               Point temp=A;                A=B;                B=temp;            }                  long double t1,t2;            Line L=Line(A,B-A);            vector<Point> inter;            getLineCircleIntersection(L, C, t1, t2,inter);            Point inter_point;            if(OnSegment(inter[0], A, B))                inter_point=inter[0];            else            {                inter_point=inter[1];            }            long double s=fabs(Cross(inter_point, A)/2);            s+=C.r*C.r*Angle(inter_point,B)/2;            return s*sg;    }}int main(){        double ld[9];    Point p[4];    while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf", &ld[0], &ld[1], &ld[2], &ld[3], &ld[4], &ld[5], &ld[6], &ld[7], &ld[8])!=EOF)    {        p[0]=Point(ld[0]-ld[6],ld[1]-ld[7]);        p[1]=Point(ld[2]-ld[6],ld[3]-ld[7]);        p[2]=Point(ld[4]-ld[6],ld[5]-ld[7]);        p[3]=p[0];                double ans=0;        if(dcmp(ld[8])!=0)        {            Circle C=Circle(Zero,ld[8]);                        for(int i=0;i<3;i++)                ans+=common_area(C, p[i], p[i+1]);                 }                printf("%.2f\n",fabs(ans)+eps);    }         return 0;}