poj 1269 Intersecting Lines

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
");
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
",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
");
63                 else printf("NONE
");
64             }
65             else if (fabs(v.y)<=exp && fabs(w.y)<=exp)
66             {
67                 if (fabs(y1-y3)<=exp)
68                 printf("LINE
");
69                 else printf("NONE
");
70             }
71             else
72             {
73                 if (fabs((y3-w.y/w.x*x3)-(y1-w.y/w.x*x1))<=exp)
74                 printf("LINE
");
75                 else printf("NONE
");
76             }
77         }
78         }
79         printf("END OF OUTPUT
");
80     }
81     return 0;
82 }