POJ 3675 Telescope（圆和多边形面积交）

题目链接：http://poj.org/problem?id=3675

题意：一个圆心在原点（0,0）的圆，给出半径和另外一个多边形P求圆和多边形的相交面积

思路：模板。。

 #include <iostream>

 #include <cstdio>

 #include <cmath>

 #define max(x,y) ((x)>(y)?(x):(y))

 #define min(x,y) ((x)<(y)?(x):(y))

 #define PI acos(-1.0)

 #define EPS 1e-10

 using namespace std;

 

 inline int DB(double x)

 {

     if(x<-EPS) return -1;

     if(x>EPS) return 1;

     return 0;

 }

 

 

 struct point

 {

     double x,y;

 

     point(){}

     point(double _x,double _y):x(_x),y(_y){}

 

     point operator-(point a)

     {

         return point(x-a.x,y-a.y);

     }

     point operator+(point a)

     {

         return point(x+a.x,y+a.y);

     }

     

     double operator*(point a)

     {

         return x*a.y-y*a.x;

     }

     

     point oppose()

     {

         return point(-x,-y);

     }

     

     double length()

     {

         return sqrt(x*x+y*y);

     }

     

     point adjust(double L)

     {

         L/=length();

         return point(x*L,y*L);

     }

     

     point vertical()

     {

         return point(-y,x);

     }

     

     double operator^(point a)

     {

         return x*a.x+y*a.y;

     }

 };

 

 

 struct segment

 {

     point a,b;

 

     segment(){}

     segment(point _a,point _b):a(_a),b(_b){}

 

     point intersect(segment s)

     {

         double s1=(s.a-a)*(s.b-a);

         double s2=(s.b-b)*(s.a-b);

         double t=s1+s2;

         s1/=t;s2/=t;

         return point(a.x*s2+b.x*s1,a.y*s2+b.y*s1);

     }

     point vertical(point p)

     {

         point t=(b-a).vertical();

         return intersect(segment(p,p+t));

     }

     int isonsegment(point p)

     {

         return DB(min(a.x,b.x)-p.x)<=0&&

                DB(max(a.x,b.x)-p.x)>=0&&

                DB(min(a.y,b.y)-p.y)<=0&&

                DB(max(a.y,b.y)-p.y)>=0;

     }

 };

 

 struct circle

 {

     point p;

     double R;

 };

 

 circle C;

 point p[105];

 int n;

 

 

 double cross_area(point a,point b,circle C)

 {

     point p=C.p;

     double R=C.R;

     int sgn=DB((b-p)*(a-p));

     double La=(a-p).length(),Lb=(b-p).length();

     int ra=DB(La-R),rb=DB(Lb-R);

     double ang=acos(((b-p)^(a-p))/(La*Lb));

     segment t(a,b);

     point s;

     point h,u,temp;

     double ans,L,d,ang1;

 

     if(!DB(La)||!DB(Lb)||!sgn) ans=0;

     else if(ra<=0&&rb<=0) ans=fabs((b-p)*(a-p))/2;

     else if(ra>=0&&rb>=0)

     {

         h=t.vertical(p);

         L=(h-p).length();

         if(!t.isonsegment(h)||DB(L-R)>=0) ans=R*R*(ang/2);

         else

         {

             ans=R*R*(ang/2);

             ang1=acos(L/R);

             ans-=R*R*ang1;

             ans+=R*sin(ang1)*L;

         }

     }

     else

     {

         h=t.vertical(p);

         L=(h-p).length();

         s=b-a;

         d=sqrt(R*R-L*L);

         s=s.adjust(d);

         if(t.isonsegment(h+s)) u=h+s;

         else u=h+s.oppose();

         if(ra==1) temp=a,a=b,b=temp;

         ans=fabs((a-p)*(u-p))/2;

         ang1=acos(((u-p)^(b-p))/((u-p).length()*(b-p).length()));

         ans+=R*R*(ang1/2);

     }

     return ans*sgn;

 }

 

 double cal_cross(circle C,point p[],int n)

 {

     double ans=0;

     int i;

     p[n]=p[0];

     for(i=0;i<n;i++) ans+=cross_area(p[i],p[i+1],C);

     return fabs(ans);

 }

 

 

 int main()

 {

     while(scanf("%lf%d",&C.R,&n)!=-1)

     {

         int i;

         for(i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);

         C.p.x=C.p.y=0;

         double area=cal_cross(C,p,n);

         printf("%.2lf\n",area);

     }

     return 0;

 }