int segIntersectCheck(const point &a, const point &b, const point &c, const point &d, point &o) {
static double s1, s2, s3, s4;
int d1 = sign(s1 = det(b - a, c - a)), d2 = sign(s2 = det(b - a, d - a));
int d3 = sign(s3 = det(d - c, a - c)), d4 = sign(s4 = det(d - c, b - c));
if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) {
o = (c * s2 - d * s1) / (s2 - s1); return true;
} iCnt = 0;
if (d1 == 0 && c.onSeg(a, b)) o = c, ++iCnt;
if (d2 == 0 && d.onSeg(a, b)) o = d, ++iCnt;
if (d3 == 0 && a.onSeg(c, d)) o = a, ++iCnt;
if (d4 == 0 && b.onSeg(c, d)) o = b, ++iCnt;
return iCnt ? 2 : 0; // `不相交返回0, 严格相交返回1, 非严格相交返回2`
}
bool segIntersect(const point &a, const point &b, const point &c, const point &d, point &e) {
double s1 = det(c - a, d - a), s2 = det(d - b, c - b);
if (!sign(s1 + s2)) return false;
e = (b - a) * (s1 / (s1 + s2)) + a;
return e.onSeg(a, b) && e.onSeg(c, d);
}
