POJ1269：直线与直线的关系判断

POJ1269

题解

• 两直线ab和cd平行：ab×cd = 0
• 两直线重合：在平行的基础上满足ac×ad = 0
• 否则求交点。

代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
using namespace std;
double const eps = 1e-12;
int n;
typedef struct Point{
	double x,y;
	Point(){};
	Point(double x,double y):x(x),y(y){};
	Point operator - (const Point& e)const{   //减
		return Point(x - e.x,y - e.y);
	}
	Point operator + (const Point& e)const{   //减
		return Point(x + e.x,y + e.y);
	}
	double operator ^ (const Point& e)const{	  //叉乘
		return x * e.y - y * e.x;
	}
}Vector;
struct Line{
	Point a,b;
}line1,line2;
int dcmp(double x){
	if(fabs(x) < eps)	return 0;
	else 	return x < 0 ? -1 : 1;
}
int Judge(Line line1,Line line2){
	if(dcmp((line1.b - line1.a) ^ (line2.b - line2.a)) == 0){    //平行
		if(dcmp((line2.a - line1.a) ^ (line2.b - line1.a)) == 0)	return 1; 	//重合  
		else	return 2;
	}else 	return 3;
}
void intersection(Point p,Vector v,Point q,Vector w){   //两直线(p+tv)和(q+tw)求交点
	Vector u = p - q;
	double t = (w ^ u) / (v ^ w);
	Point ans = p + Point(v.x * t,v.y * t);
	printf("POINT %.2f %.2f\n",ans.x,ans.y);
}
int main(){
	scanf("%d",&n);
	printf("INTERSECTING LINES OUTPUT\n");
	for(int i=1;i<=n;i++){
		scanf("%lf%lf%lf%lf",&line1.a.x,&line1.a.y,&line1.b.x,&line1.b.y);
		scanf("%lf%lf%lf%lf",&line2.a.x,&line2.a.y,&line2.b.x,&line2.b.y);
		int flag = Judge(line1,line2);
		if(flag == 1)	printf("LINE\n");
		else if(flag == 2)	printf("NONE\n");
		else intersection(line1.a,line1.b-line1.a,line2.a,line2.b-line2.a);
	}
	printf("END OF OUTPUT\n");
	return 0;
}