You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7568    Accepted Submission(s): 3691

Many geometry（几何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very
easy, after all we are now attending an exam, not a contest
Give you N (1<=N<=100) segments（线段）, please output the number of all intersections（交点）. 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.

2
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.00
3
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.000
0.00 0.00 1.00 0.00
0

1
3

/*
hdoj 1086
就是线段相交的个数 
*/
#include<iostream>
#include<stdio.h>
using namespace std;

struct point{    
    double x,y;    
};        
   
point point1[101],point2[101];
 
double xmult(point a,point b,point c)//大于零代表a,b,c左转    
{      
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);      
}      
    
bool OnSegment(point a,point b,point c)//a,b,c共线时有效     
{              
    return c.x>=min(a.x,b.x)&&c.x<=max(a.x,b.x)&&c.y>=min(a.y,b.y)&&c.y<=max(a.y,b.y);        
}      
      
int cross(point a,point b,point c,point d){//判断ab 与cd是否相交       
    double d1,d2,d3,d4;      
    d1=xmult(c,d,a);      
    d2=xmult(c,d,b);      
    d3=xmult(a,b,c);      
    d4=xmult(a,b,d);      
    if(d1*d2<0&&d3*d4<0)  return 1;      
    else    if(d1==0&&OnSegment(c,d,a)) return 1;      
    else    if(d2==0&&OnSegment(c,d,b)) return 1;      
    else    if(d3==0&&OnSegment(a,b,c)) return 1;      
    else    if(d4==0&&OnSegment(a,b,d)) return 1;      
    return 0;      
}       
int main()
{
	int n,i,j,sum,f;
	
//	freopen("test.txt","r",stdin);
	while(scanf("%d",&n)&&n)	
	{
		for(i=0;i<n;i++)
			scanf("%lf%lf%lf%lf",&point1[i].x,&point1[i].y,&point2[i].x,&point2[i].y);
		sum=0;
		for(i=0;i<n-1;i++)
			for(j=i+1;j<n;j++)
			{
				f=cross(point1[i],point2[i],point1[j],point2[j]);//2条线段 
				if(f==1) sum++;
			}
		printf("%d\n",sum);
	}
	return 0;
}