poj 1269 计算几何

 1 /**
 2 判断直线位置关系
 3 **/
 4 #include <iostream>
 5 #include <cmath>
 6 #include <cstdio>
 7 using namespace std;
 8 struct point {
 9     double x,y;
10     point(double x=0,double y=0):x(x),y(y){}
11 };
12 
13 typedef point Vector;
14 
15 Vector operator - (point A,point B){
16     return Vector(A.x-B.x,A.y-B.y);
17 }
18 
19 struct line {
20     point a,b;
21 };
22 double length(Vector v){
23     return sqrt(v.x*v.x+v.y*v.y);
24 }
25 
26 double  cross(Vector A,Vector B){
27     return A.x*B.y-A.y*B.x;
28 }
29 
30 double distoline(point P,point A,point B){
31     Vector v1 =B-A,v2 = P-A;
32     return fabs(cross(v1,v2))/length(v1);
33 }
34 
35 int main()
36 {
37     int n;
38     cin>>n;
39     line l1,l2;
40     cout<<"INTERSECTING LINES OUTPUT"<<endl;
41     while(n--){
42         cin>>l1.a.x>>l1.a.y>>l1.b.x>>l1.b.y;
43         cin>>l2.a.x>>l2.a.y>>l2.b.x>>l2.b.y;
44         Vector tmp1,tmp2;
45         tmp1.x = l1.b.x-l1.a.x;
46         tmp1.y = l1.b.y-l1.a.y;
47         tmp2.x = l2.b.x-l2.a.x;
48         tmp2.y = l2.b.y-l2.a.y;
49         //cout<<tmp1.x<<" "<<tmp1.y<<endl;
50         //cout<<tmp2.x<<" "<<tmp2.y<<endl;
51         //cout<<cross(tmp1,tmp2)<<endl;
52         if(cross(tmp1,tmp2)==0){
53             if(distoline(l1.a,l2.a,l2.b)==0){
54                 cout<<"LINE"<<endl;
55             }else
56                 cout<<"NONE"<<endl;
57         }else{
58             double x,y;
59             if(l1.a.x==l1.b.x&&l2.a.x!=l2.b.x){
60                 x = l1.a.x;
61                 double k = (l2.b.y-l2.a.y)/(l2.b.x-l2.a.x);
62                 double b = l2.a.y-k*l2.a.x;;
63                 y = k*x+b;
64             }else if(l1.a.x!=l1.b.x&&l2.a.x==l2.b.x){
65                 x = l2.a.x;
66                 double k = (l1.b.y-l1.a.y)/(l1.b.x-l1.a.x);
67                 double b = l1.a.y-k*l1.a.x;
68                 y = k*x+b;
69             }else{
70                 double k1= (l1.b.y-l1.a.y)/(l1.b.x-l1.a.x);
71                 double b1=l1.a.y-k1*l1.a.x;
72                 double k2 = (l2.b.y-l2.a.y)/(l2.b.x-l2.a.x);
73                 double b2=l2.a.y-k2*l2.a.x;
74                 x =(b2-b1)/(k1-k2);
75                 y = k1*x+b1;
76             }
77 
78             printf("POINT %.2lf %.2lf\n",x,y);
79         }
80     }
81     cout<<"END OF OUTPUT"<<endl;
82     return 0;
83 }