poj 3335 Rotating Scoreboard - 半平面交
/*
poj 3335 Rotating Scoreboard - 半平面交
点是顺时针给出的
*/
#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);
}
//求多边形面积
double mo_area_polygon(point *dian,int n)
{
int i;
point yuan;
yuan.x=yuan.y=0;
double ret=0;
for(i=0;i<n;++i)
{
ret+=mo_xmult(dian[(i+1)%n],yuan,dian[i]);
}
return ret;
}
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];
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_ge(a,0))//逆时针就是<=0
{
ans[k++]=ans[i];
}else if(mo_ll(a*b,0))
{
ans[k++]=mo_intersection(qian,hou,ans[i],ans[(i+1)%nofdian]);
}
}
nofdian=k;
}
int mo_banjiao(point *dian,int n)
{
int i,nofdian;
double area=mo_area_polygon(dian,n);
if(area<0)
{
point temp;
int zhong=n/2-1;
for(i=0;i<=zhong;++i)
{
temp=dian[i];
dian[i]=dian[n-1-i];
dian[n-1-i]=temp;
}
}
nofdian=n;
for(i=0;i<n;++i)
{
mo_banjiao_jiao[i]=dian[i];
}
for(i=1;i<n;++i)
{
mo_banjiao_cut(mo_banjiao_jiao,dian[i],dian[(i+1)%n],nofdian);
}
return nofdian;
}
/////////////////////////
int main()
{
int t,i,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&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("NO\n");
}else
{
printf("YES\n");
}
}
return 0;
}