首先判断是否相交，就是枚举3*3对边的相交关系。
如果不相交，判断包含还是相离，就是判断点在三角形内还是三角形外。两边各判断一次。
//http://acm.fzu.edu.cn/problem.php?pid=2273
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std; const double eps=1e-;
const double pi=acos(-1.0);
int sgn(double x)
{
if (fabs(x)<eps) return ;
if (x<) return -;
return ;
}
struct Point
{
double x,y;
Point() {}
Point(double _x,double _y)
{
x=_x;
y=_y;
}
Point operator +(const Point &b) const
{
return Point(x+b.x,y+b.x);
}
Point operator -(const Point &b) const
{
// return Point(x-b.x,y-b.x);
return Point(x-b.x,y-b.y);
}
double operator ^(const Point &b) const
{
return x*b.y-y*b.x;
}
double operator *(const Point &b) const
{
return x*b.x+y*b.y;
}
Point operator /(const double b) const
{
return Point(x/b,y/b);
}
}; struct Line
{
Point s,e;
Line() {}
Line(Point _s,Point _e)
{
s=_s;
e=_e;
}
}; double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
bool inter(Line l1,Line l2)
{
return
max(l1.s.x,l1.e.x)>=min(l2.s.x,l2.e.x) &&
max(l2.s.x,l2.e.x)>=min(l1.s.x,l1.e.x) &&
max(l1.s.y,l1.e.y)>=min(l2.s.y,l2.e.y) &&
max(l2.s.y,l2.e.y)>=min(l1.s.y,l1.e.y) &&
sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e))<= &&
sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e))<=;
}
Point o;
bool _cmp(Point p1,Point p2)
{
double tmp=(p1-o)^(p2-o);
if (sgn(tmp)>) return true;
else if (sgn(tmp)== && sgn(dist(p1,o)-dist(p2,o))<=) return true;
else return false;
} bool OnSeg(Point P,Line L)
{
return
sgn((L.s-P)^(L.e-P))== &&
sgn((P.x-L.s.x)*(P.x-L.e.x))<= &&
sgn((P.y-L.s.y)*(P.y-L.e.y))<=;
}
int inConvexPoly(Point a,Point p[],int n)
{
for (int i=;i<n;i++)
{
if (sgn((p[i]-a)^(p[(i+)%n]-a))<) return -; // out
else if (OnSeg(a,Line(p[i],p[(i+)%n]))) return ; // on
}
return ; // in
} Point p[]; int main()
{
int t;
scanf("%d",&t);
while (t--)
{
for (int i=;i<;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
bool xj=false;
for (int i=;i<;i++)
{
for (int j=;j<;j++)
{
Line l1=Line(p[i],p[(i+)%]);
Line l2=Line(p[j+],p[(j+)%+]);
if (inter(l1,l2))
{
xj=true;
break;
}
}
if (xj) break;
}
if (xj)
{
printf("intersect\n");
continue;
}
int in1=,in2=;
o=(p[]+p[]+p[])/3.0;
sort(p,p+,_cmp);
o=(p[]+p[]+p[])/3.0;
sort(p+,p+,_cmp);
for (int i=;i<;i++)
{
if (inConvexPoly(p[i],p+,)==) in1++;
if (inConvexPoly(p[i+],p,)==) in2++;
}
if (in1==||in2==) printf("contain\n");
else printf("disjoint\n");
}
return ;
}