poj 1269 Intersecting Lines

人生第一道计算几何

Description

We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect.
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.
Input

The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).
Output

There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read “END OF OUTPUT”.
Sample Input

5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5
Sample Output

INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT

这道题也是对以后的预告吧。。

判断直线关系

平行：两条直线各任选两个点组成两个向量平行（叉积为0）

相交：只需要判断是否平行即可（叉积为0 ）

重合：在平行的基础上，在两条直线上各选一个点组成一个向量在去与前两个判平行（叉积为0）

直线交点

设直线分别为\(P+tV\)和\(Q+tW\)且设向量\(u=P-Q\)，设交点在直线1的参数为\(t_1\)，交点在直线2的参数为\(t_2\)

这样可以列一个方程，解得

\(t_1=\frac{cross(w,u)}{cross(v,w)},t_2=\frac{cross(v,u)}{cross(v,w)}\)

然后直接搞就行了

代码：

#include <cstdio>
#include <cmath>
#include <cstdlib>
using namespace std;

const double eps=1e-9;
int dcmp(double x)
{
	if (fabs(x)<eps) return 0;
	else
		if (x>0)
			return 1;
		else
			return -1;
}

const double pi=acos(-1.0);

struct Vector
{
	double x,y;
	Vector (double a=0,double b=0)
	{
		x=a,y=b;
	}
};
typedef Vector Point;

struct Line
{
	Point p;
	Vector v;
	Line(Point P=Point(0,0),Vector V=Vector(0,0))
	{
		p=P,v=V;
	}
};

Vector operator + (Vector a,Vector b) {return Vector(a.x+b.x,a.y+b.y);}
Vector operator - (Vector a,Vector b) {return Vector(a.x-b.x,a.y-b.y);}
Vector operator * (Vector a,double b) {return Vector(a.x*b  ,a.y*b  );}
Vector operator / (Vector a,double b) {return Vector(a.x/b  ,a.y/b  );}
bool   operator ==(Vector a,Vector b) {return !dcmp(a.x-b.x)&&!dcmp(a.y-b.y);}

Vector Rotate(Vector a,double rad)
{
	return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}

double Dot(Vector a,Vector b)
{
    return a.x*b.x+a.y*b.y;
}

double Len(Vector a){return sqrt(Dot(a,a));}

double Angle(Vector a,Vector b){return acos(Dot(a,b)/Len(a)/Len(b));}

double Cross(Vector a,Vector b)
{
    return a.x*b.y-a.y*b.x;
}

Vector Normal(Vector a)
{
    double L=Len(a);
    return Vector(-a.y/L,a.x/L);
}

Point Line_cross(Line a,Line b)
{
	Vector u=a.p-b.p;
	double t=Cross(b.v,u)/Cross(a.v,b.v);
	return a.p+a.v*t;
}

main()
{
	int T;
	double x1,x2,x3,x4,y1,y2,y3,y4;
	puts("INTERSECTING LINES OUTPUT");
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
		Line l1,l2;
		l1=Line(Point(x1,y1),Vector(x2-x1,y2-y1));
		l2=Line(Point(x3,y3),Vector(x4-x3,y4-y3));
		if (!dcmp(Cross(l1.v,l2.v))&&!dcmp(Cross(l1.v,l1.p-l2.p)))
			puts("LINE");
		else if (!dcmp(Cross(l1.v,l2.v)))
			puts("NONE");
		else
		{
			Point ans=Line_cross(l1,l2);
			printf("POINT %.2lf %.2lf\n",ans.x,ans.y);
		}
	}
	puts("END OF OUTPUT");
}