POJ 1269 Intersecting Lines(线段相交,水题)
大意:给你两条直线的坐标,推断两条直线是否共线、平行、相交。若相交。求出交点。
思路:线段相交推断、求交点的水题。没什么好说的。
struct Point{
double x, y;
} ;
struct Line{
Point a, b;
} A, B;
double xmult(Point p1, Point p2, Point p)
{
return (p1.x-p.x)*(p2.y-p.y)-(p1.y-p.y)*(p2.x-p.x);
}
bool parallel(Line u, Line v)
{
return zero((u.a.x-u.b.x)*(v.a.y-v.b.y)-(v.a.x-v.b.x)*(u.a.y-u.b.y));
}
Point intersection(Line u, Line v)
{
Point ret = u.a;
double t = ((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
ret.x += (u.b.x-u.a.x)*t, ret.y += (u.b.y-u.a.y)*t;
return ret;
}
int T;
void Solve()
{
scanf("%d", &T);
printf("INTERSECTING LINES OUTPUT\n");
while(T--)
{
scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &A.a.x, &A.a.y, &A.b.x, &A.b.y, &B.a.x, &B.a.y, &B.b.x, &B.b.y);
if(parallel(A, B) && zero(xmult(A.a, B.a, B.b)))
{
printf("LINE\n");
}
else if(parallel(A, B))
{
printf("NONE\n");
}
else
{
Point t = intersection(A, B);
printf("POINT %.2f %.2f\n", t.x, t.y);
}
}
printf("END OF OUTPUT\n");
}