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-8;
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 -1;//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 0;//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 1;
}
int main(){
    int n;
    Point p1,p2,p3,p4;
    scanf("%d",&n);
    printf("INTERSECTING LINES OUTPUT
");  
    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==-1 ) puts("LINE");  
        else if( f==0 ) puts("NONE");  
        else{
            printf("POINT %.2lf %.2lf
",P_ans.x,P_ans.y);
        }
    }
    printf("END OF OUTPUT
");  
    return 0;
}