POJ 1269 Intersecting Lines（判断两条直线的位置关系）

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

题意：判断两条直线的位置关系，平行，重合，如果相交，并求出交点。

如果两个向量的叉积为0，那么这两个向量平行（共线），再进一步判断如果一条直线上的一个点在令一条直线上，那么这两条直线重合，否者平行。

要求两直线的交点，通过变量t将直线p1-p2上的点p0表示为p1+t(p2-p1)，交点又在p4-p3上，所以有

(p4-p3)X(p1+t(p2-p1)-p3) = 0

于是p0 = p1 + ((p4-p3)X(p3-p1))/((p4-p3)X(p2-p1))*(p2-p1）

#include<cmath>#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const double eps = 1e-6;struct Point{    double x,y;    Point(){}    Point(double _x,double _y)    {        x = _x;y = _y;    }    Point operator -(const Point &b)const    {        return Point(x - b.x,y - b.y);    }    double operator *(const Point &b)const    {        return x*b.x + y*b.y;    }    double operator ^(const Point &b)const    {        return x*b.y - y*b.x;    }};int main(void){    int T,i,j,k;    scanf("%d",&T);    printf("INTERSECTING LINES OUTPUT\n");    while(T--)    {        double x1,y1,x2,y2,x3,y3,x4,y4;        scanf("%lf%lf%lf%lf %lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);        Point p1(x1,y1),p2(x2,y2),p3(x3,y3),p4(x4,y4);        if(fabs((p2-p1)^(p4-p3)) < eps)        {            if(fabs((p3-p1)^(p4-p1)) < eps)                printf("LINE\n");            else                printf("NONE\n");        }        else        {            double t = ((p4-p3)^(p3-p1))/((p4-p3)^(p2-p1));            double x = p1.x + (p2.x-p1.x)*t;            double y = p1.y + (p2.y-p1.y)*t;            printf("POINT %.2f %.2f\n",x,y);        }    }    printf("END OF OUTPUT\n");    return 0;}