題面：

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 OUTPUTPOINT 2.00 2.00NONELINEPOINT 2.00 5.00POINT 1.07 2.20END OF OUTPUT思路：本題求的就是兩條直線之間的位置關係，如果平行輸出“NONE”，相交輸出“POINT”和交點座標，重合就輸出“LINE”。判斷兩條直線是否平行則判斷兩條直線的單位方向向量是否相等或相反（即斜率是否相等），如果滿足則是平行或重合，否則就是相交，相交就呼叫求交點的函式求出交點即可；而判斷是否重合只需判斷一條直線上的某一點是否在另一條直線上即可。
程式碼實現如下：

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

struct Point {
    double x, y;
    Point (double x = 0, double y = 0) : x(x), y(y) {}
};

typedef Point Vector;

int n;
Point A, B, C, D;

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 p) {
    return Vector(A.x * p, A.y * p);
}

bool operator < (const Point& a, const Point& b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

const double eps = 1e-10;
int dcmp(double x) {
    if(fabs(x) < eps)
        return 0;
    else
        return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point& b) {
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector A, Vector B) {
    return A.x * B.x + A.y * B.y;
}

double Length(Vector A) {
    return sqrt(Dot(A, A));
}

double Cross(Vector A, Vector B) {
    return A.x * B.y - A.y * B.x;
}

//求單位方向向量
Vector Unit_direction_vector(Vector w) {
    return Vector(w.x / Length(w), w.y / Length(w));
}

//判斷兩直線是否不相交
bool isIntersection(Vector A, Vector B) {
    return Unit_direction_vector(A) == Unit_direction_vector(B) || Unit_direction_vector(Vector(- A.x, - A.y)) == Unit_direction_vector(B);
}

Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {
    Vector u = P - Q;
    double t = Cross (w, u) / Cross(v, w);
    return P + v * t;
}

//判斷兩直線是否重合只要判斷是否有公共點即可
bool OnLine(Point p, Point a1, Point a2) {
    return dcmp(Cross(a1 - p, a2 - p)) == 0;
}


int main() {
    while(~scanf("%d", &n)) {
        printf("INTERSECTING LINES OUTPUT\n");
        while(n--) {
            scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y, &D.x, &D.y);
            if(isIntersection(A - B, C - D)) {
                if(OnLine(A, C, D)) {
                    printf("LINE\n");
                } else {
                    printf("NONE\n");
                }
            } else {
                Point P = GetLineIntersection(A, A - B, C, C - D);
                printf("POINT %.2f %.2f\n", P.x, P.y);
            }
        }
        printf("END OF OUTPUT\n");
    }
}