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

 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;

 #define eps 1e-8

 struct P{
     double x,y;

     P(){}
     P(double x,double y):x(x),y(y){}

     friend P operator +(P a,P b){
         return P(a.x+b.x,a.y+b.y);
     }
     friend P operator -(P a,P b){
         return P(a.x-b.x,a.y-b.y);
     }
     friend double operator *(P a,P b){
         return a.x * b.y - a.y * b.x ;
     }
     friend double operator /(P a,P b){
         return a.x * b.x + a.y * b.y ;
     }
     friend bool operator ==(P a,P b){
         return fabs(a.x-b.x)<eps && fabs(a.y-b.y)<eps;
     }
     friend bool operator !=(P a,P b){
         return !(a==b);
     }
     friend bool operator <(P a,P b){
         if (fabs(a.y-b.y)<eps) return a.x<b.x;
         return a.y<b.y;
     }
     friend double turn(P a,P b,P c){    //向量AB X 向量AC
         return (b-a)*(c-a);
     }
     friend void print(P a){
         printf("%.2lf %.2lf\n",a.x,a.y);
     }
 };

 struct L{
     P a,b;
     L(){}
     L(P a,P b):a(a),b(b){}
     friend P subl(L a){  //向量l
         return P(a.b.x-a.a.x,a.b.y-a.a.y);
     }
 };

 bool online(L l,P p){
     if (fabs(turn(l.a,l.b,p))<eps) return ;
     return ;
 }
 P solve(double a1,double b1,double c1,double a2,double b2,double c2){
     P a;
     a.x = (b1*c2-b2*c1)/(a1*b2-a2*b1);
     a.y = (a1*c2-a2*c1)/(a2*b1-a1*b2);
     return a;
 }
 void check(L lx,L ly){
     if (online(lx,ly.a) && online(lx,ly.b)) printf("LINE\n"); else
     if (fabs(subl(lx)*subl(ly))<eps) printf("NONE\n");     else
     {
         double x1=lx.a.x , y1=lx.a.y;
         double x2=lx.b.x , y2=lx.b.y;
         double x3=ly.a.x , y3=ly.a.y;
         double x4=ly.b.x , y4=ly.b.y;
         double a1 = y1 - y2 , b1 = x2 - x1 , c1 = x1*y2 - x2*y1 ;
         double a2 = y3 - y4 , b2 = x4 - x3 , c2 = x3*y4 - x4*y3 ;
         printf("POINT ");
         print(solve(a1,b1,c1,a2,b2,c2));
     }
 }

 int main(){
     int T;
     scanf("%d",&T);
     printf("INTERSECTING LINES OUTPUT\n");
     while (T--){
         double x1,y1,x2,y2,x3,y3,x4,y4;
         scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
         P a(x1,y1),b(x2,y2),c(x3,y3),d(x4,y4);
         L lx(a,b),ly(c,d);
         check(lx,ly);
     }
     printf("END OF OUTPUT\n");
 }