POJ1269+直线相交

求相交点

 1 /*

 2 线段相交模板：判相交、求交点

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #include<math.h>

 8 

 9 const double eps = 1e-8;

10 struct Point{

11     double x,y;

12 };

13 Point P_ans;

14 double cross( Point a,Point b,Point c ){

15     return ( b.x-a.x )*( c.y-a.y )-( b.y-a.y )*( c.x-a.x );

16 }

17 int solve( Point a,Point b,Point c,Point d ){

18     if( fabs(cross(a,b,c))<=eps&&fabs(cross(a,b,d))<=eps )

19         return -1;//两条线段在同一条直线上

20     if( fabs((b.x-a.x)*(d.y-c.y)-(b.y-a.y)*(d.x-c.x))<=eps )

21         return 0;//两条线断平行

22     /*

23     求交点用到叉积(必须保证有交点)

24     交点为p0(x,y)

25     (A-p0)*(B-p0)=0

26     (C-p0)*(D-p0)=0

27     */

28     double a1,a2,b1,b2,c1,c2;

29     a1 = a.y-b.y,b1 = b.x-a.x,c1 = a.x*b.y-b.x*a.y;

30     a2 = c.y-d.y,b2 = d.x-c.x,c2 = c.x*d.y-d.x*c.y;

31     P_ans.x = (b1*c2-b2*c1)/(a1*b2-a2*b1); 

32     P_ans.y = (a2*c1-a1*c2)/(a1*b2-a2*b1); 

33     return 1;

34 }

35 int main(){

36     int n;

37     Point p1,p2,p3,p4;

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

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

40     while( n-- ){

41         scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y,&p3.x,&p3.y,&p4.x,&p4.y);

42         int f = solve( p1,p2,p3,p4 );

43         if( f==-1 ) puts("LINE");  

44         else if( f==0 ) puts("NONE");  

45         else{

46             printf("POINT %.2lf %.2lf\n",P_ans.x,P_ans.y);

47         }

48     }

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

50     return 0;

51 }

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