POJ 1269 Intersecting Lines（线段相交，水题）

大意：给你两条直线的坐标，判断两条直线是否共线、平行、相交，若相交，求出交点。

思路：线段相交判断、求交点的水题，没什么好说的。

 1 struct Point{
 2     double x, y;
 3 } ;
 4 struct Line{
 5     Point a, b;
 6 } A, B;
 7 
 8 double xmult(Point p1, Point p2, Point p)
 9 {
10     return (p1.x-p.x)*(p2.y-p.y)-(p1.y-p.y)*(p2.x-p.x);
11 }
12 
13 bool parallel(Line u, Line v)
14 {
15     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));
16 }
17 
18 Point intersection(Line u, Line v)
19 {
20     Point ret = u.a;
21     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));
22     ret.x += (u.b.x-u.a.x)*t, ret.y += (u.b.y-u.a.y)*t;
23     return ret;
24 }
25 
26 int T;
27 
28 void Solve()
29 {
30     scanf("%d", &T);
31     printf("INTERSECTING LINES OUTPUT
");
32     while(T--)
33     {
34         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);
35         if(parallel(A, B) && zero(xmult(A.a, B.a, B.b)))
36         {
37             printf("LINE
");
38         }
39         else if(parallel(A, B))
40         {
41             printf("NONE
");
42         }
43         else
44         {
45             Point t = intersection(A, B);
46             printf("POINT %.2f %.2f
", t.x, t.y);
47         }
48     }
49     printf("END OF OUTPUT
");
50 }

POJ 1269