POJ 1269 (直线相交) Intersecting Lines

水题，以前总结的模板还是很好用的。

 1 #include <cstdio>

 2 #include <cmath>

 3 using namespace std;

 4 

 5 const double eps = 1e-8;

 6 

 7 int dcmp(double x)

 8 {

 9     if(fabs(x) < eps) return 0;

10     return x < 0 ? -1 : 1;

11 }

12 

13 struct Point

14 {

15     double x, y;

16     Point(double x=0, double y=0):x(x), y(y) {}

17 };

18 typedef Point Vector;

19 

20 Point read_point()

21 {

22     double x, y;

23     scanf("%lf%lf", &x, &y);

24     return Point(x, y);

25 }

26 

27 Point operator + (const Point& A, const Point& B)

28 { return Point(A.x+B.x, A.y+B.y); }

29 

30 Point operator - (const Point& A, const Point& B)

31 { return Point(A.x-B.x, A.y-B.y); }

32 

33 Point operator * (const Point& A, double p)

34 { return Point(A.x*p, A.y*p); }

35 

36 double Cross(const Point& A, const Point& B)

37 { return A.x*B.y - A.y*B.x; }

38 

39 double Length(const Vector v)

40 { return sqrt(v.x*v.x + v.y*v.y); }

41 

42 double DistanceToLine(Point P, Point A, Point B)

43 {

44     Vector v1 = B - A, v2 = P - A;

45     return fabs(Cross(v1, v2)) / Length(v1);

46 }

47 

48 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w)

49 {

50     Vector u = P - Q;

51     double t = Cross(w, u) / Cross(v, w);

52     return P + v*t;

53 }

54 

55 int main()

56 {

57     //freopen("in.txt", "r", stdin);

58 

59     int n;

60     scanf("%d", &n);

61     puts("INTERSECTING LINES OUTPUT");

62     Point A, B, C, D;

63     while(n--)

64     {

65         A = read_point(); B = read_point();

66         C = read_point(); D = read_point();

67         Vector v1 = B - A, v2 = D - C;

68         if(dcmp(Cross(v1, v2)) == 0)

69         {

70             if(dcmp(DistanceToLine(C, A, B)) == 0) puts("LINE");

71             else puts("NONE");

72 

73         }

74         else

75         {

76             Point ans = GetLineIntersection(A, v1, C, v2);

77             printf("POINT %.2f %.2f\n", ans.x, ans.y);

78         }

79     }

80     puts("END OF OUTPUT");

81 

82     return 0;

83 }

代码君