三角剖分·圆和多边形的交

POJ 3675 Telescope

http://poj.org/problem?id=3675

大意：求解圆和多边形的交。
分析：任意一个凸N多边形均可分解成N-2个三角形。因此，这就是讨论分解后的三角形和圆的交的问题。
它有这些情况:
（1）：



（2）：



（3）:



（4）:



(5):



5)又可分为p1在外，p2在里；p1在里，p2在外。
在写代码的过程中遇到了一些无法解决的问题，特别是5.2情况，参考了
http://gzhu-101majia.iteye.com/blog/1128345
不能理解为什么自己的做法不能通过。
求解p1p2直线和圆的交点（假设不与横轴垂直，可参照(5)图）：



分解后的三角形和圆的交按有向面积累加：

#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;
const double eps=1e-7,PI=acos(-1.0);
struct point
{
double x,y;
}p[55];
double r;

double pp_dis(point p1,point p2) //两点距离
{
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double xmulti(point p0,point p1,point p2){
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
double angle(point p1,point o,point p2){
double a=pp_dis(o,p1);
double b=pp_dis(o,p2);
double c=pp_dis(p1,p2);
double cos=(a*a+b*b-c*c)/(2*a*b);
return fabs(acos(cos));
}
double dir(point o,point p1,point p2){
double area=xmulti(o,p1,p2);
if(area<-eps) return 1.0;
return -1.0;
}
point intersec(double x,double y)    //计算直线与圆的交点
{
double k;
point temp;
if(x!=0)
{
k=y/x;
temp.x=fabs(r)/sqrt(1+k*k);
if(x<0) temp.x=-temp.x;
temp.y=k*temp.x;
}
else
{
temp.x=0;
if(y>0) temp.y=r;
else temp.y=-r;
}
return temp;
}
double get_area(point o,point p1,point p2) //三角剖分
{
double f=dir(o,p1,p2);    //判断三角形面积加还是减
double s;
double a=pp_dis(o,p1);
double b=pp_dis(o,p2);
double c=pp_dis(p1,p2);
double d=fabs(xmulti(o,p1,p2))/c;
// 1
if(a<=r && b<=r)
{
double area=xmulti(o,p1,p2);
return fabs(area)/2.0*f;
}
// 2
else if(a>=r && b>=r && d>=r)
{
double sita1=angle(p1,o,p2);
double s=fabs(sita1*r*r/2.0); //扇形s=θ*r*r/2
return s*f;
}

// 3
else if(a>=r && b>=r && d<=r && (angle(o,p1,p2)-PI/2>eps || angle(o,p2,p1)-PI/2>eps))
{
double sita=angle(p1,o,p2);
s=fabs(sita*r*r/2.0);
return s*f;
}
// 4
else if(a>=r && b>=r && d<=r && angle(o,p1,p2)<=PI/2 && angle(o,p2,p1)<=PI/2)
{
point p3,p4;
if(fabs(p1.x-p2.x)>eps){
double k=(p2.y-p1.y)/(p2.x-p1.x);
double A=1+k*k;
double B=2*p1.y*k-2*k*k*p1.x;
double C=k*k*p1.x*p1.x+p1.y*p1.y-2*p1.y*k*p1.x-r*r;
double x1=(-B-sqrt(B*B-4*A*C))/(2*A);
double x2=(-B+sqrt(B*B-4*A*C))/(2*A);
if(fabs(x1-p1.x)<fabs(x2-p1.x)){
p3.x=x1;
p4.x=x2;
}
else {
p3.x=x2;
p4.x=x1;
}
p3.y=p1.y+k*(p3.x-p1.x);
p4.y=p1.y+k*(p4.x-p1.x);
}
else {
p3.x=p1.x;
p4.x=p1.x;
double h=sqrt(r*r-d*d);
if(p1.y>eps){
p3.y=h;
p4.y=-h;
}
else {
p3.y=-h;
p4.y=h;
}
}

double ag1=angle(p1,o,p3);
double ag2=angle(p4,o,p2);
double area=0;
area+=r*r*(ag1+ag2)/2;
area+=fabs(xmulti(p3,o,p4))/2.0;
return area*f;
}

// 5.1
else if(a>=r && b<=r)
{
double area=0;
point p3;
if(fabs(p1.x-p2.x)>eps){
double k=(p2.y-p1.y)/(p2.x-p1.x);
double A=1+k*k;
double B=-2*p1.x*k*k+2*p1.y*k;
double C=p1.y*p1.y+k*k*p1.x*p1.x-2*p1.y*k*p1.x-r*r;
double x1=(-B+sqrt(B*B-4*A*C))/(2*A);
double x2=(-B-sqrt(B*B-4*A*C))/(2*A);
if( (x1>p1.x&&x1<p2.x) || (x1>p2.x&&x1<p1.x)) p3.x=x1;
else p3.x=x2;
p3.y=p1.y+k*(p3.x-p1.x);
}
else {
p3.x=p1.x;
double y1=sqrt(r*r-p3.x*p3.x);
double y2=-y1;
if( (y1>p1.y&&y1<p2.y) || (y1>p2.y&&y1<p1.y)) p3.y=y1;
else p3.y=y2;
}
area+=fabs(xmulti(p3,o,p2))/2.0;
double ag=angle(p3,o,p1);
area+=r*r*ag/2;
return area*f;
}
//  5.2
else if(a<=r && b>=r)    //a短于半径，b长于半径
{
double area=0;
point p3;
if(fabs(p1.x-p2.x)>eps)
{
double k=(p2.y-p1.y)/(p2.x-p1.x);
double A=1+k*k;
double B=-2*p1.x*k*k+2*p1.y*k;
double C=p1.y*p1.y+k*k*p1.x*p1.x-2*p1.y*k*p1.x-r*r;
double x1=(-B+sqrt(B*B-4*A*C))/(2*A);
double x2=(-B-sqrt(B*B-4*A*C))/(2*A);
if( (x1>p1.x&&x1<p2.x) || (x1>p2.x&&x1<p1.x)) p3.x=x1;
else p3.x=x2;
p3.y=p1.y+k*(p3.x-p1.x);
}
else
{
p3.x=p1.x;
double y1=sqrt(r*r-p3.x*p3.x);
double y2=-y1;
if( (y1>p1.y&&y1<p2.y) || (y1>p2.y&&y1<p1.y)) p3.y=y1;
else p3.y=y2;
}
//double sita1=angle(p2,o,p3);
point pp2=intersec(p2.x,p2.y);
double c=pp_dis(pp2,p3);
double cos=(r*r+r*r-c*c)/(2*r*r);
double sita1=acos(cos);  //让我想不通的是为什么求角度非得用这种方法，而上面更加直接的方法
double s1=fabs(sita1*r*r/2.0); // 却错误。
double s3=fabs(p3.x*p1.y-p3.y*p1.x)/2.0;
area=s1+s3;
return area*f;
}
else return 0;
}

int main()
{
//freopen("cin.txt","r",stdin);
while(~scanf("%lf",&r)){
double ans=0;
point o;
o.x=o.y=0;
int n;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%lf%lf",&p[i].x,&p[i].y);
}
p
=p[0];
ans=0;
for(int i=0;i<n;i++)
{
ans+=get_area(o,p[i],p[i+1]);
}
printf("%.2lf\n",fabs(ans));
}
return 0;
}

POJ 2986 A Triangle and a Circle

http://poj.org/problem?id=2986

求解三角形和圆的交。圆心是可变的。但是三角形只有一个。

再一次用此方法，但是看似更加简单的这题居然过不了，无语啊，太打击人的积极性了吧！！
WA：

#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;
const double eps=1e-7,PI=acos(-1.0);
struct point{
double x,y;
};
double pp_dis(point p1,point p2) //两点距离
{
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double xmulti(point p0,point p1,point p2){
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
double angle(point p1,point o,point p2){
double a=pp_dis(o,p1);
double b=pp_dis(o,p2);
double c=pp_dis(p1,p2);
double cos=(a*a+b*b-c*c)/(2*a*b);
return fabs(acos(cos));
}
double dir(point o,point p1,point p2){
double area=xmulti(o,p1,p2);
if(area<-eps) return 1.0;
return -1.0;
}
point intersec(double r,double x,double y)    //计算直线与圆的交点
{
double k;
point temp;
if(x!=0)
{
k=y/x;
temp.x=fabs(r)/sqrt(1+k*k);
if(x<0) temp.x=-temp.x;
temp.y=k*temp.x;
}
else
{
temp.x=0;
if(y>0) temp.y=r;
else temp.y=-r;
}
return temp;
}
double get_area(double r,point o,point p1,point p2) //三角剖分
{
double f=dir(o,p1,p2);    //判断三角形面积加还是减
double s;
double a=pp_dis(o,p1);
double b=pp_dis(o,p2);
double c=pp_dis(p1,p2);
double d=fabs(xmulti(o,p1,p2))/c;
// 1
if(a<=r && b<=r)
{
double area=xmulti(o,p1,p2);
return fabs(area)/2.0*f;
}
// 2
else if(a>=r && b>=r && d>=r)
{
double sita1=angle(p1,o,p2);
double s=fabs(sita1*r*r/2.0); //扇形s=θ*r*r/2
return s*f;
}

// 3
else if(a>=r && b>=r && d<=r && (angle(o,p1,p2)-PI/2>eps || angle(o,p2,p1)-PI/2>eps))
{
double sita=angle(p1,o,p2);
s=fabs(sita*r*r/2.0);
return s*f;
}
// 4
else if(a>=r && b>=r && d<=r && angle(o,p1,p2)<=PI/2 && angle(o,p2,p1)<=PI/2)
{
point p3,p4;
if(fabs(p1.x-p2.x)>eps){
double k=(p2.y-p1.y)/(p2.x-p1.x);
double A=1+k*k;
double B=2*p1.y*k-2*k*k*p1.x;
double C=k*k*p1.x*p1.x+p1.y*p1.y-2*p1.y*k*p1.x-r*r;
double x1=(-B-sqrt(B*B-4*A*C))/(2*A);
double x2=(-B+sqrt(B*B-4*A*C))/(2*A);
if(fabs(x1-p1.x)<fabs(x2-p1.x)){
p3.x=x1;
p4.x=x2;
}
else {
p3.x=x2;
p4.x=x1;
}
p3.y=p1.y+k*(p3.x-p1.x);
p4.y=p1.y+k*(p4.x-p1.x);
}
else {
p3.x=p1.x;
p4.x=p1.x;
double h=sqrt(r*r-d*d);
if(p1.y>eps){
p3.y=h;
p4.y=-h;
}
else {
p3.y=-h;
p4.y=h;
}
}

double ag1=angle(p1,o,p3);
double ag2=angle(p4,o,p2);
double area=0;
area+=r*r*(ag1+ag2)/2;
area+=fabs(xmulti(p3,o,p4))/2.0;
return area*f;
}
// 5.1
else if(a>=r && b<=r)
{
double area=0;
point p3;
if(fabs(p1.x-p2.x)>eps){
double k=(p2.y-p1.y)/(p2.x-p1.x);
double A=1+k*k;
double B=-2*p1.x*k*k+2*p1.y*k;
double C=p1.y*p1.y+k*k*p1.x*p1.x-2*p1.y*k*p1.x-r*r;
double x1=(-B+sqrt(B*B-4*A*C))/(2*A);
double x2=(-B-sqrt(B*B-4*A*C))/(2*A);
if( (x1>p1.x&&x1<p2.x) || (x1>p2.x&&x1<p1.x)) p3.x=x1;
else p3.x=x2;
p3.y=p1.y+k*(p3.x-p1.x);
}
else {
p3.x=p1.x;
double y1=sqrt(r*r-p3.x*p3.x);
double y2=-y1;
if( (y1>p1.y&&y1<p2.y) || (y1>p2.y&&y1<p1.y)) p3.y=y1;
else p3.y=y2;
}
area+=fabs(xmulti(p3,o,p2))/2.0;
double ag=angle(p3,o,p1);
area+=r*r*ag/2;
return area*f;
}
//  5.2
else if(a<=r && b>=r)    //a短于半径，b长于半径
{
double area=0;
point p3;
if(fabs(p1.x-p2.x)>eps)
{
double k=(p2.y-p1.y)/(p2.x-p1.x);
double A=1+k*k;
double B=-2*p1.x*k*k+2*p1.y*k;
double C=p1.y*p1.y+k*k*p1.x*p1.x-2*p1.y*k*p1.x-r*r;
double x1=(-B+sqrt(B*B-4*A*C))/(2*A);
double x2=(-B-sqrt(B*B-4*A*C))/(2*A);
if( (x1>p1.x&&x1<p2.x) || (x1>p2.x&&x1<p1.x)) p3.x=x1;
else p3.x=x2;
p3.y=p1.y+k*(p3.x-p1.x);
}
else
{
p3.x=p1.x;
double y1=sqrt(r*r-p3.x*p3.x);
double y2=-y1;
if( (y1>p1.y&&y1<p2.y) || (y1>p2.y&&y1<p1.y)) p3.y=y1;
else p3.y=y2;
}
//double sita1=angle(p2,o,p3);
point pp2=intersec(r,p2.x,p2.y);
double c=pp_dis(pp2,p3);
double cos=(r*r+r*r-c*c)/(2*r*r);
double sita1=acos(cos);
double s1=fabs(sita1*r*r/2.0);
double s3=fabs(xmulti(o,p1,p3))/2.0;
area=s1+s3;
return area*f;
}
else return 0;
}

int main()
{
//freopen("cin.txt","r",stdin);
point p[5],o;
double r;
while(~scanf("%lf",&p[1].x)){
scanf("%lf",&p[1].y);
for(int i=2;i<=3;i++){
scanf("%lf %lf",&p[i].x,&p[i].y);
}
scanf("%lf %lf %lf",&o.x,&o.y,&r);
for(int i=1;i<=3;i++){
p[i].x-=o.x;
p[i].y-=o.y;
}
p[4]=p[1];
o.x=0;  o.y=0;
double ans=0;
for(int i=1;i<=3;i++){
ans=ans+get_area(r,o,p[i],p[i+1]);
}
//printf("%.2f\n",fabs(ans));
printf("%.2f\n",fabs(ans)+eps);
}
return 0;
}

当我使用模板后,主函数内这样写：

Point p[5],o;
double r;
while(~scanf("%lf",&p[1].x)){
scanf("%lf",&p[1].y);
for(int i=2;i<=3;i++){
scanf("%lf %lf",&p[i].x,&p[i].y);
}
scanf("%lf %lf %lf",&o.x,&o.y,&r);
for(int i=1;i<=3;i++){
p[i].x-=o.x;
p[i].y-=o.y;
}
p[4]=p[1];

double ans=0;
Circle C=Circle(Zero,r);
for(int i=1;i<=3;i++)
ans+=common_area(C, p[i], p[i+1]);
printf("%.2lf\n",fabs(ans)+eps);
}
return 0;

输出0. 不能直接给x,y赋值，对于long double数据成员的点%Lf输入居然也输出0。只能利用构造函数来初始化了
主函数内：

Point p[5];
double a[9];
while(~scanf("%lf",&a[0])){
for(int i=1;i<9;i++){
scanf("%lf",&a[i]);
}
p[0]=Point(a[0]-a[6],a[1]-a[7]);
p[1]=Point(a[2]-a[6],a[3]-a[7]);
p[2]=Point(a[4]-a[6],a[5]-a[7]);
p[3]=p[0];

double ans=0;
Circle C=Circle(Zero,a[8]);
for(int i=0;i<3;i++)
ans+=common_area(C, p[i], p[i+1]);
printf("%.2lf\n",fabs(ans)+eps);
}

代码：

#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()
{
//freopen("cin.txt","r",stdin);
Point p[5];
double a[9];
while(~scanf("%lf",&a[0])){
for(int i=1;i<9;i++){
scanf("%lf",&a[i]);
}
p[0]=Point(a[0]-a[6],a[1]-a[7]);
p[1]=Point(a[2]-a[6],a[3]-a[7]);
p[2]=Point(a[4]-a[6],a[5]-a[7]);
p[3]=p[0];

double ans=0;
Circle C=Circle(Zero,a[8]);
for(int i=0;i<3;i++)
ans+=common_area(C, p[i], p[i+1]);
printf("%.2lf\n",fabs(ans)+eps); // C++ AC G++ TLE
}
return 0;
}