HDU 1086 You can Solve a Geometry Problem too（计算几何）

Description
给出n条线段的端点坐标，问这n条线段的交点个数，保证不会出现多线共点的情况
Input
多组输入，每组用例第一行为线段条数n，之后n行每行四个浮点数表示这n条线段两段点的横纵坐标，以n=0结束输入
Output
对于每组用例，输出这n条线段的交点个数
Sample Input
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
Sample Output
1
3
Solution
简单计算几何，枚举任意两条线段判断是否相交即可
Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const double eps=1e-8;
#define maxn 111
int sgn(double x)
{
    if(fabs(x)<eps)return 0;
    if(x<0)return -1;
    else return 1;
}
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.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;
    }
};
struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e)
    {
        s=_s;e=_e;
    }
};
//判断线段相交
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.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s))<=0&&
        sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s))<=0;
}
int main()
{
    int n;
    Line l[maxn];
    while(scanf("%d",&n),n)
    {
        for(int i=1;i<=n;i++)
            scanf("%lf%lf%lf%lf",&l[i].s.x,&l[i].s.y,&l[i].e.x,&l[i].e.y);
        int ans=0;
        for(int i=1;i<=n;i++)
            for(int j=i+1;j<=n;j++)
                ans+=inter(l[i],l[j])?1:0;
        printf("%d\n",ans);
    }
    return 0;
}