题目链接：
题面：
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
50 0 4 4 0 4 4 05 0 7 6 1 0 2 35 0 7 6 3 -6 4 -32 0 2 27 1 5 18 50 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：本题求的就是两条直线之间的位置关系，如果平行输出“NONE”，相交输出“POINT”和交点坐标，重合就输出“LINE”。判断两条直线是否平行则判断两条直线的单位方向向量是否相等或相反（即斜率是否相等），如果满足则是平行或重合，否则就是相交，相交就调用求交点的函数求出交点即可；而判断是否重合只需判断一条直线上的某一点是否在另一条直线上即可。 代码实现如下：
1 #include2 #include 3 #include 4 using namespace std; 5 6 struct Point { 7 double x, y; 8 Point (double x = 0, double y = 0) : x(x), y(y) {} 9 };10 11 typedef Point Vector;12 13 int n;14 Point A, B, C, D;15 16 Vector operator + (Vector A, Vector B) {17 return Vector(A.x + B.x, A.y + B.y);18 }19 20 Vector operator - (Vector A, Vector B) {21 return Vector(A.x - B.x, A.y - B.y);22 }23 24 Vector operator * (Vector A, double p) {25 return Vector(A.x * p, A.y * p);26 }27 28 bool operator < (const Point& a, const Point& b) {29 return a.x < b.x || (a.x == b.x && a.y < b.y);30 }31 32 const double eps = 1e-10;33 int dcmp(double x) {34 if(fabs(x) < eps)35 return 0;36 else37 return x < 0 ? -1 : 1;38 }39 40 bool operator == (const Point& a, const Point& b) {41 return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;42 }43 44 double Dot(Vector A, Vector B) {45 return A.x * B.x + A.y * B.y;46 }47 48 double Length(Vector A) {49 return sqrt(Dot(A, A));50 }51 52 double Cross(Vector A, Vector B) {53 return A.x * B.y - A.y * B.x;54 }55 56 //求单位方向向量57 Vector Unit_direction_vector(Vector w) {58 return Vector(w.x / Length(w), w.y / Length(w));59 }60 61 //判断两直线是否不相交62 bool isIntersection(Vector A, Vector B) {63 return Unit_direction_vector(A) == Unit_direction_vector(B) || Unit_direction_vector(Vector(- A.x, - A.y)) == Unit_direction_vector(B);64 }65 66 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {67 Vector u = P - Q;68 double t = Cross (w, u) / Cross(v, w);69 return P + v * t;70 }71 72 //判断两直线是否重合只要判断是否有公共点即可73 bool OnLine(Point p, Point a1, Point a2) {74 return dcmp(Cross(a1 - p, a2 - p)) == 0;75 }76 77 78 int main() {79 while(~scanf("%d", &n)) {80 printf("INTERSECTING LINES OUTPUT\n");81 while(n--) {82 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);83 if(isIntersection(A - B, C - D)) {84 if(OnLine(A, C, D)) {85 printf("LINE\n");86 } else {87 printf("NONE\n");88 }89 } else {90 Point P = GetLineIntersection(A, A - B, C, C - D);91 printf("POINT %.2f %.2f\n", P.x, P.y);92 }93 }94 printf("END OF OUTPUT\n");95 }96 }