Give you N (1<=N<=100) segments(線段), please output the number of all intersections(交點). You should count repeatedly if M (M>2) segments intersect at the same point.
Note:
You can assume that two segments would not intersect at more than one point.
A test case starting with 0 terminates the input and this test case is not to be processed.
#include <cstdio>
#include <cmath>
const int maxn=105;
struct Point{
double x,y;
Point(double x=0,double y=0):x(x),y(y){}
}A[maxn],B[maxn];
typedef Point Vector;
Vector operator-(Point A,Point B){return Vector(A.x-B.x,A.y-B.y);}
const double eps=1e-10;
int dcmp(double x){
if(fabs(x)<eps) return 0;else return x<0?-1:1;
}
double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}
bool Segments_X(Point a1,Point a2,Point b1,Point b2){
double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),
c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2)<=0&&dcmp(c3)*dcmp(c4)<=0;
}
int main(){
int n;
while(scanf("%d",&n)==1&&n){
for(int i=0;i<n;i++) scanf("%lf%lf%lf%lf",&A[i].x,&A[i].y,&B[i].x,&B[i].y);
int ans=0;
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++)
if(Segments_X(A[i],B[i],A[j],B[j])) ans++;
printf("%d\n",ans);
}
return 0;
}