/*
poj 1474 Video Surveillance - 求多边形有没有核
*/
#include <stdio.h>
#include<math.h>
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<eps) return 1;
return 0;
}
inline bool mo_gg(double x,double y) { return x > y + eps;} // x > y
inline bool mo_ll(double x,double y) { return x < y - eps;} // x < y
inline bool mo_ge(double x,double y) { return x > y - eps;} // x >= y
inline bool mo_le(double x,double y) { return x < y + eps;} // x <= y
inline 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<nofdian;++i)
{
double a,b;
a=mo_xmult(hou,ans[i],qian);
b=mo_xmult(hou,ans[(i+1)%nofdian],qian);
if(mo_le(a,0))//顺时针就是<=0
{
mo_banjiao_jiao_temp[k++]=ans[i];
}if(mo_ll(a*b,0))
{
mo_banjiao_jiao_temp[k++]=mo_intersection(qian,hou,ans[i],ans[(i+1)%nofdian]);
}
}
for(i=0;i<k;++i)
{
ans[i]=mo_banjiao_jiao_temp[i];
}
nofdian=k;
}
int mo_banjiao(point *dian,int n)
{
int i,nofdian;
nofdian=n;
for(i=0;i<n;++i)
{
mo_banjiao_jiao[i]=dian[i];
}
for(i=0;i<n;++i)//i从0开始
{
mo_banjiao_cut(mo_banjiao_jiao,dian[i],dian[(i+1)%n],nofdian);
if(nofdian==0)
{
return nofdian;
}
}
return nofdian;
}
/////////////////////////
int main()
{
int t,i,n,iofcase=1;
while(scanf("%d",&n),n)
{
for(i=0;i<n;++i)
{
scanf("%lf%lf",&dian[i].x,&dian[i].y);
}
int ret=mo_banjiao(dian,n);
if(ret==0)
{
printf("Floor #%d
",iofcase++);
printf("Surveillance is impossible.
");
}else
{
printf("Floor #%d
",iofcase++);
printf("Surveillance is possible.
");
}
}
return 0;
}