POJ1269+直线相交

求相交点

 /*
 线段相交模板：判相交、求交点
 */
 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #include<math.h>

 const double eps = 1e-;
 struct Point{
     double x,y;
 };
 Point P_ans;
 double cross( Point a,Point b,Point c ){
     return ( b.x-a.x )*( c.y-a.y )-( b.y-a.y )*( c.x-a.x );
 }
 int solve( Point a,Point b,Point c,Point d ){
     if( fabs(cross(a,b,c))<=eps&&fabs(cross(a,b,d))<=eps )
         return -;//两条线段在同一条直线上
     if( fabs((b.x-a.x)*(d.y-c.y)-(b.y-a.y)*(d.x-c.x))<=eps )
         return ;//两条线断平行
     /*
     求交点用到叉积(必须保证有交点)
     交点为p0(x,y)
     (A-p0)*(B-p0)=0
     (C-p0)*(D-p0)=0
     */
     double a1,a2,b1,b2,c1,c2;
     a1 = a.y-b.y,b1 = b.x-a.x,c1 = a.x*b.y-b.x*a.y;
     a2 = c.y-d.y,b2 = d.x-c.x,c2 = c.x*d.y-d.x*c.y;
     P_ans.x = (b1*c2-b2*c1)/(a1*b2-a2*b1);
     P_ans.y = (a2*c1-a1*c2)/(a1*b2-a2*b1);
     return ;
 }
 int main(){
     int n;
     Point p1,p2,p3,p4;
     scanf("%d",&n);
     printf("INTERSECTING LINES OUTPUT\n");
     while( n-- ){
         scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y,&p3.x,&p3.y,&p4.x,&p4.y);
         int f = solve( p1,p2,p3,p4 );
         if( f==- ) puts("LINE");
         else if( f== ) puts("NONE");
         else{
             printf("POINT %.2lf %.2lf\n",P_ans.x,P_ans.y);
         }
     }
     printf("END OF OUTPUT\n");
     return ;
 }