【題目】

傳送門

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

【分析】

大致題意：給出兩條直線，判斷這兩條直線的關係（重合，平行或相交），若相交還要輸出交點

題解：平面幾何入門題

以下用 $cross$ 表示叉積，用 $p_1,p_2,p_3,p_4$ 表示四個點， $p_1,p_2$ 表示 $l_1$， $p_3,p_4$ 表示 $l_2$

1. 若兩直線重合，則 $p_3$ 應該在 $l_1$ 上，且 $p_4$ 也應該在 $l_1$ 上，用叉積判斷一下即可
2. 若兩直線平行，則兩直線的叉積應為 $0$（可以理解成形成的平行四邊形面積為 $0$）
3. 若上述兩種情況都不符合，那就肯定相交，求交點可以用代數法（即求 $k$ 和 $b$），也可以用幾何法，這裡就不多說怎麼實現了，不懂的就上網搜吧

【程式碼】

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define eps 1e-8
using namespace std;
struct point
{
	double x,y;
	point(){}
	point(double x,double y):x(x),y(y){}
	point operator+(const point &a)  {return point(x+a.x,y+a.y);}
	point operator-(const point &a)  {return point(x-a.x,y-a.y);}
	double operator*(const point &a)  {return x*a.x+y*a.y;}
	double operator^(const point &a)  {return x*a.y-a.x*y;}
};
struct line
{
	point start,end;
}a,b;
point mul(point a,double k)
{
	return point(a.x*k,a.y*k);
}
bool Collinear(line a,line b)
{
	double s1=(a.end-a.start)^(b.start-a.start);
	double s2=(a.end-a.start)^(b.end-a.start);
	return fabs(s1)<eps&&fabs(s2)<eps;
}
bool Parallel(line a,line b)
{
	double s1=(a.end-a.start)^(b.end-b.start);
	return fabs(s1)<eps;
}
point inter(line p,line q)
{
	double s1=(p.start-q.start)^(p.end-q.start);
	double s2=(p.start-q.end)^(p.end-q.end);
	double k=s1/(s1-s2);
	return mul(q.end-q.start,k)+q.start;
}
int main()
{
	int n,i;
	scanf("%d",&n);
	int x1,y1,x2,y2,x3,y3,x4,y4;
	printf("INTERSECTING LINES OUTPUT\n");
	for(i=1;i<=n;++i)
	{
		scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
		scanf("%d%d%d%d",&x3,&y3,&x4,&y4);
		a.start=point(x1,y1),a.end=point(x2,y2);
		b.start=point(x3,y3),b.end=point(x4,y4);
		if(Collinear(a,b))  printf("LINE\n");
		else  if(Parallel(a,b))  printf("NONE\n");
		else
		{
			point p=inter(a,b);
			printf("POINT %.2f %.2f\n",p.x,p.y);
		}
	}
	printf("END OF OUTPUT");
	return 0;
}