【模板】判断线段是否相交

如果相交返回true，不相交返回false

bool check(Node a, Node b, Node c, Node d){
	if(!(min(a.x,b.x) <= max(c.x,d.x)
	 && min(c.y,d.y) <= max(a.y,b.y) 
	 && min(c.x,d.x) <= max(a.x,b.x)
	 && min(a.y,b.y) <= max(c.y,d.y))) 
	     return 0;//所在矩形都不重叠 肯定不相交
    double u, v, w, z;
    u = (c.x - a.x) * (b.y - a.y) - (b.x - a.x) * (c.y - a.y);
	v = (d.x - a.x) * (b.y - a.y) - (b.x - a.x) * (d.y - a.y);
	w = (a.x - c.x) * (d.y - c.y) - (d.x - c.x) * (a.y - c.y);	
	z = (b.x - c.x) * (d.y - c.y) - (d.x - c.x) * (b.y - c.y);
    return u * v < -eps && w * z < -eps;//跨立实验 解释如下 
}

关于跨立实验的解释(转自勿忘初心0924)

如果两条线段相交，那么必须跨立，就是以一条线段为标准，另一条线段的两端点一定在这条线段的两段
也就是说a b两点在线段cd的两端，c d两点在线段ab的两端
这里就用到了向量X乘的知识点，有向量X乘的物理意义知：AB x CD=-CD x AB
看下图：

(ca x cd)·(cb x cd)<=0 则说明ca cb先对于cd的方向不同，则a b在线段cd的两侧，由此可以判断其他点