/*poj 1474 Video Surveillance - 求多边形有没有核*/#include 
    
     #include
     
      const double eps=1e-8;const int N=103;struct point{    double x,y;}dian[N];inline bool mo_ee(double x,double y){	double ret=x-y;	if(ret<0) ret=-ret;	if(ret
      
        y + eps;} // x > yinline bool mo_ll(double x,double y)  {   return x < y - eps;} // x < yinline bool mo_ge(double x,double y) {   return x > y - eps;} // x >= yinline bool mo_le(double x,double y) {   return x < y + eps;} 	// x <= yinline double mo_xmult(point p2,point p0,point p1)//p1在p2左返回负，在右边返回正{    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}point mo_intersection(point u1,point u2,point v1,point v2){    point ret=u1;    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))		/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));    ret.x+=(u2.x-u1.x)*t;    ret.y+=(u2.y-u1.y)*t;    return ret;}///切割法求半平面交point mo_banjiao_jiao[N*2];point mo_banjiao_jiao_temp[N*2];void mo_banjiao_cut(point *ans,point qian,point hou,int &nofdian){	int i,k;	for(i=k=0;i