POJ 1269 Intersecting Lines【判断直线相交】

#include<stdio.h>
#include<string.h>
#include<math.h>
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{//点积 
        return x*b.x+y*b.y;
    }
    double operator ^(const point &b)const{//叉积 
        return x*b.y-y*b.x;
    }
}A,B,C,D;
double cal(point p0,point p1,point p2){//小于0表示在p1处左折，大于0右折，等于0同线 
    return (p1-p0)^(p2-p0);
}
const double eps=1e-8;
int main(){
    int n,i;
    while(scanf("%d",&n)!=EOF){
        puts("INTERSECTING LINES OUTPUT");
        for(i=1;i<=n;i++){
            scanf("%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y);
            scanf("%lf%lf%lf%lf",&C.x,&C.y,&D.x,&D.y);
            if(fabs(cal(C,A,B))<eps&&fabs(cal(D,A,B))<eps)    puts("LINE");
            else if((B.x-A.x)*(D.y-C.y)==(D.x-C.x)*(B.y-A.y))    puts("NONE");
            else{
                double a1=A.y-B.y;
                double b1=B.x-A.x;
                double c1=A.x*B.y-B.x*A.y;
                double a2=C.y-D.y;
                double b2=D.x-C.x;
                double c2=C.x*D.y-D.x*C.y;
                double x=(b1*c2-b2*c1)/(a1*b2-a2*b1);
                double y=(a2*c1-a1*c2)/(a1*b2-a2*b1);
                printf("POINT %.2f %.2f\n",x,y);
            }
        }
        puts("END OF OUTPUT");
    }
    return 0;
}