poj 1269 Intersecting Lines

题目链接：http://poj.org/problem?id=1269

题目大意：给出四个点的坐标x1,y1,x2,y2,x3,y3,x4,y4,前两个形成一条直线，后两个坐标形成一条直线。然后问你是否平行，重叠或者相交，如果相交，求出交点坐标。

算法：二维几何直线相交+叉积

解法：先用叉积判断是否相交，如果相交的话，设交点坐标为p0(x0,y0)。向量（p0p1）和(p0p2)的叉积为0，有(x1-x0)*(y2-y0)-(y1-y0)*(x2-x0)=0;同理，求出p0和p3p4直线的式子。然后联立求解x0,y0。平行或重叠的情况就自己YY了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #define inf 0x7fffffff
 #define exp 1e-10
 #define PI 3.141592654
 using namespace std;
 struct Point
 {
     double x,y;
     Point(double x=,double y=):x(x),y() {}
 };
 typedef Point Vector;
 double cross(Vector A,Vector B)
 {
     return A.x*B.y-A.y*B.x;
 }
 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
 {
     Point uu;
     Vector u=Point(P.x-Q.x , P.y-Q.y);
     double t=cross(w,u)/cross(v,w);
     uu.x=P.x+v.x*t;
     uu.y=P.y+v.y*t;
     return uu;
 }//调用训练指南上这个函数怎么错了，我写错了吗
 int main()
 {
     int n;
     double x1,y1,x2,y2,x3,y3,x4,y4;
     //cin>>n;
     while (cin>>n)
     {
         printf("INTERSECTING LINES OUTPUT\n");
         while (n--) {
         cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
         Point P,Q;
         P.x=x1 ;P.y=y1 ;
         Q.x=x3 ;Q.y=y3 ;
         Vector v,w;
         v.x=x2-x1 ;v.y=y2-y1;
         w.x=x4-x3 ;w.y=y4-y3;
         if (cross(v,w)!=)
         {
             //Vector vv=GetLineIntersection(P,v,Q,w);
             double a1,b1,c1;
             double a2,b2,c2;
             a1=y1-y2 ;b1=x2-x1 ;c1=x1*y2-y1*x2;
             a2=y3-y4 ;b2=x4-x3 ;c2=x3*y4-y3*x4;
             double x0=(b1*c2-b2*c1)/(b2*a1-b1*a2);
             double y0=(a2*c1-a1*c2)/(a1*b2-a2*b1);
             printf("POINT %.2f %.2f\n",x0,y0);
         }
         else
         {
             if (fabs(v.x)<=exp && fabs(w.x)<=exp)
             {
                 if (fabs(x1-x3)<=exp)
                 printf("LINE\n");
                 else printf("NONE\n");
             }
             else if (fabs(v.y)<=exp && fabs(w.y)<=exp)
             {
                 if (fabs(y1-y3)<=exp)
                 printf("LINE\n");
                 else printf("NONE\n");
             }
             else
             {
                 if (fabs((y3-w.y/w.x*x3)-(y1-w.y/w.x*x1))<=exp)
                 printf("LINE\n");
                 else printf("NONE\n");
             }
         }
         }
         printf("END OF OUTPUT\n");
     }
     return ;
 }