poj 1269 Intersecting Lines

题目链接：http://poj.org/problem?id=1269

题目大意：给出四个点的坐标x1,y1,x2,y2,x3,y3,x4,y4,前两个形成一条直线，后两个坐标形成一条直线。然后问你是否平行，重叠或者相交，如果相交，求出交点坐标。

算法：二维几何直线相交+叉积

解法：先用叉积判断是否相交，如果相交的话，设交点坐标为p0(x0,y0)。向量（p0p1）和(p0p2)的叉积为0，有(x1-x0)*(y2-y0)-(y1-y0)*(x2-x0)=0;同理，求出p0和p3p4直线的式子。然后联立求解x0,y0。平行或重叠的情况就自己YY了。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<cstdlib>

 5 #include<cmath>

 6 #include<algorithm>

 7 #define inf 0x7fffffff

 8 #define exp 1e-10

 9 #define PI 3.141592654

10 using namespace std;

11 struct Point

12 {

13     double x,y;

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

15 };

16 typedef Point Vector;

17 double cross(Vector A,Vector B)

18 {

19     return A.x*B.y-A.y*B.x;

20 }

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

22 {

23     Point uu;

24     Vector u=Point(P.x-Q.x , P.y-Q.y);

25     double t=cross(w,u)/cross(v,w);

26     uu.x=P.x+v.x*t;

27     uu.y=P.y+v.y*t;

28     return uu;

29 }//调用训练指南上这个函数怎么错了，我写错了吗

30 int main()

31 {

32     int n;

33     double x1,y1,x2,y2,x3,y3,x4,y4;

34     //cin>>n;

35     while (cin>>n)

36     {

37         printf("INTERSECTING LINES OUTPUT\n");

38         while (n--) {

39         cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;

40         Point P,Q;

41         P.x=x1 ;P.y=y1 ;

42         Q.x=x3 ;Q.y=y3 ;

43         Vector v,w;

44         v.x=x2-x1 ;v.y=y2-y1;

45         w.x=x4-x3 ;w.y=y4-y3;

46         if (cross(v,w)!=0)

47         {

48             //Vector vv=GetLineIntersection(P,v,Q,w);

49             double a1,b1,c1;

50             double a2,b2,c2;

51             a1=y1-y2 ;b1=x2-x1 ;c1=x1*y2-y1*x2;

52             a2=y3-y4 ;b2=x4-x3 ;c2=x3*y4-y3*x4;

53             double x0=(b1*c2-b2*c1)/(b2*a1-b1*a2);

54             double y0=(a2*c1-a1*c2)/(a1*b2-a2*b1);

55             printf("POINT %.2f %.2f\n",x0,y0);

56         }

57         else

58         {

59             if (fabs(v.x)<=exp && fabs(w.x)<=exp)

60             {

61                 if (fabs(x1-x3)<=exp)

62                 printf("LINE\n");

63                 else printf("NONE\n");

64             }

65             else if (fabs(v.y)<=exp && fabs(w.y)<=exp)

66             {

67                 if (fabs(y1-y3)<=exp)

68                 printf("LINE\n");

69                 else printf("NONE\n");

70             }

71             else

72             {

73                 if (fabs((y3-w.y/w.x*x3)-(y1-w.y/w.x*x1))<=exp)

74                 printf("LINE\n");

75                 else printf("NONE\n");

76             }

77         }

78         }

79         printf("END OF OUTPUT\n");

80     }

81     return 0;

82 }