int cross(point &a,point &b,point &c)
{
int x1=b.x-a.x;
int y1=b.y-a.y;
int x2=c.x-a.x;
int y2=c.y-a.y;
return x1*y2-x2*y1;
}
向量叉积(图解)
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转载自blog.csdn.net/sm20170867238/article/details/88374221
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