POJ 1269 (直线求交）

Problem Intersecting Lines (POJ 1269)

题目大意

　　给定两条直线，问两条直线是否重合，是否平行，或求出交点。

解题分析

　　主要用叉积做，可以避免斜率被0除的情况。

　　求交点P0: 已知P1 P2 P3 P4

　　运用 P0P1 X P0P2 = 0 和 P0P3 X P0P4 = 0 

　　C++ 用%.2lf g++ 用 %.2f!!!

　　C++ 用%.2lf g++ 用 %.2f!!!

　　C++ 用%.2lf g++ 用 %.2f!!!

参考程序

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 using namespace std;
 6 
 7 #define eps 1e-8
 8 
 9 struct P{
10     double x,y;
11     
12     P(){}
13     P(double x,double y):x(x),y(y){}
14 
15     friend P operator +(P a,P b){
16         return P(a.x+b.x,a.y+b.y);
17     } 
18     friend P operator -(P a,P b){
19         return P(a.x-b.x,a.y-b.y);
20     } 
21     friend double operator *(P a,P b){
22         return a.x * b.y - a.y * b.x ;
23     }
24     friend double operator /(P a,P b){
25         return a.x * b.x + a.y * b.y ;
26     }
27     friend bool operator ==(P a,P b){
28         return fabs(a.x-b.x)eps;
29     }
30     friend bool operator !=(P a,P b){
31         return !(a==b);
32     }
33     friend bool operator <(P a,P b){
34         if (fabs(a.y-b.y)return a.x<b.x;
35         return a.y<b.y;
36     }
37     friend double turn(P a,P b,P c){    //向量AB X 向量AC
38         return (b-a)*(c-a);
39     }
40     friend void print(P a){
41         printf("%.2lf %.2lf\n",a.x,a.y);
42     }
43 };
44 
45 struct L{
46     P a,b;
47     L(){}
48     L(P a,P b):a(a),b(b){}
49     friend P subl(L a){  //向量l
50         return P(a.b.x-a.a.x,a.b.y-a.a.y);
51     }
52 };
53 
54 bool online(L l,P p){
55     if (fabs(turn(l.a,l.b,p))return 1;
56     return 0;
57 }
58 P solve(double a1,double b1,double c1,double a2,double b2,double c2){
59     P a;
60     a.x = (b1*c2-b2*c1)/(a1*b2-a2*b1);
61     a.y = (a1*c2-a2*c1)/(a2*b1-a1*b2);
62     return a;
63 }
64 void check(L lx,L ly){
65     if (online(lx,ly.a) && online(lx,ly.b)) printf("LINE\n"); else
66     if (fabs(subl(lx)*subl(ly))"NONE\n");     else
67     {
68         double x1=lx.a.x , y1=lx.a.y;
69         double x2=lx.b.x , y2=lx.b.y;
70         double x3=ly.a.x , y3=ly.a.y;
71         double x4=ly.b.x , y4=ly.b.y;
72         double a1 = y1 - y2 , b1 = x2 - x1 , c1 = x1*y2 - x2*y1 ;
73         double a2 = y3 - y4 , b2 = x4 - x3 , c2 = x3*y4 - x4*y3 ;
74         printf("POINT ");
75         print(solve(a1,b1,c1,a2,b2,c2));
76     }
77 }
78 
79 int main(){
80     int T;
81     scanf("%d",&T);
82     printf("INTERSECTING LINES OUTPUT\n");
83     while (T--){
84         double x1,y1,x2,y2,x3,y3,x4,y4;
85         scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
86         P a(x1,y1),b(x2,y2),c(x3,y3),d(x4,y4);
87         L lx(a,b),ly(c,d);
88         check(lx,ly);
89     }
90     printf("END OF OUTPUT\n");
91 }

View Code

 

转载于:https://www.cnblogs.com/rpSebastian/p/5716837.html