/*poj3348 Cows 凸包+多边形面积 水题floor向下取整，返回的是double*/#include
    
     #include
     
      #include 
      
        using namespace std;const double eps = 1e-8;  struct point{	double x,y;};int n;point dian[10000+10],zhan[10000+10];//point *mo_dian;double mo_distance(point p1,point p2){    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}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);}bool mo_ee(double x,double y){	double ret=x-y;	if(ret<0) ret=-ret;	if(ret
       
         y + eps;} // x > y   bool mo_ll(double x,double y)  {   return x < y - eps;} // x < y   bool mo_ge(double x,double y) {   return x > y - eps;} // x >= y   bool mo_le(double x,double y) {   return x < y + eps;}     // x <= y   bool mo_cmp(point a,point b)   // 第一次排序   {      if( mo_ee(a.y ,b.y ) )          return mo_ll(a.x, b.x);      return mo_ll(a.y,b.y);  }  bool mo_cmp1(point a,point b)  // 第二次排序   {      double len = mo_xmult(b,mo_dian[0],a);      if( mo_ee(len,0.0) )          return mo_ll(mo_distance(mo_dian[0],a),mo_distance(mo_dian[0],b));      return mo_gg(len,0.0);  }  int mo_graham(int n,point *dian,point *stk){	int i,top=1;	mo_dian=dian;	sort(mo_dian,mo_dian+n,mo_cmp);      sort(mo_dian+1,mo_dian+n,mo_cmp1);  	stk[0]=mo_dian[0];      stk[1]=mo_dian[1];  	for(i=2;i
        
         0&&mo_xmult(mo_dian[i],stk[top-1],stk[top])<=0) --top;          stk[++top]=mo_dian[i];      }      return top+1;  }double mo_area_polygon(point *dian,int n){	int i;	point yuan;	yuan.x=yuan.y=0;	double ret=0;	for(i=0;i