模板——判断两条线段是否相交(严格与不严格)

不包括端点和部分重合

const double eps=1e-8;

double xmult(point p1,point p2,point p0)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

int opposite_side(point p1,point p2,point l1,point l2)
{
    return xmult(l1,p1,l2)*xmult(l1,p2,l2)<-eps;
}

int intersect_ex(point u1,point u2,point v1,point v2)
{
    return opposite_side(u1,u2,v1,v2)&&opposite_side(v1,v2,u1,u2);
}
//调用intersect_ex,有交点返回1

 

包括端点和部分重合

const double eps=1e-8;
#define zero(x)(((x)>0?(x):(-x))<eps)

double xmult(point p1,point p2,point p0)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

int dots_inline(point p1,point p2,point p3)
{
    return zero(xmult(p1,p2,p3));
}

int same_side(point p1,point p2,point l1,point l2)
{
    return xmult(l1,p1,l2)*xmult(l1,p2,l2)>eps;
}

int dot_online_in(point p,point l1,point l2)
{
    return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;
}

int intersect_in(point u1,point u2,point v1,point v2)
{
    if(!dots_inline(u1,u2,v1)||!dots_inline(u1,u2,v2)) return !same_side(u1,u2,v1,v2)&&!same_side(v1,v2,u1,u2);
    return dot_online_in(u1,v1,v2)||dot_online_in(u2,v1,v2)||dot_online_in(v1,u1,u2)||dot_online_in(v2,u1,u2);
}

//调用intersect_in,相交返回1

 

posted @ 2016-01-23 09:57  Fighting_Heart  阅读(339)  评论(0编辑  收藏  举报