Intersecting Lines - POJ 1269（判断平面上两条直线的关系）

分析：有三种关系，共线，平行，还有相交，共线和平行都可以使用叉积来进行判断（其实和斜率一样），相交需要解方程....在纸上比划比划就出来了....

代码如下：

======================================================================================================================================

#include<math.h>
#include<algorithm>
#include<stdio.h>
using namespace std;

const int MAXN = ;
const double EPS = 1e-;

struct point
{///定义点
    double x, y;

    point(double x=, double y=):x(x),y(y){}
    point operator - (const point &t) const{
        return point(x-t.x, y-t.y);
    }
    double operator * (const point &t) const{
        return x * t.y - y * t.x;
    }
};
struct segment
{
    point A, B;
    double a, b, c;

    void InIt(point t1, point t2)
    {
        A = t1, B = t2;
        a = B.y - A.y;
        b = A.x - B.x;
        c =  A.x*B.y-B.x*A.y;
    }
};

bool Parallel(segment t1, segment t2)
{///是否平行
    return fabs((t1.A-t1.B)*(t2.A-t2.B)) < EPS;
}
bool Collineation(segment t1, segment t2)
{///是否共线

    return fabs((t1.A-t1.B)*(t2.A-t1.B)) < EPS  && fabs((t1.A-t1.B)*(t2.B-t1.B)) < EPS;
}
point CrossPoint(segment t1, segment t2)
{
    point t;

    t.x = (t1.c*t2.b-t2.c*t1.b) / (t1.a*t2.b-t2.a*t1.b);
    t.y = (t1.c*t2.a-t2.c*t1.a) / (t1.b*t2.a-t2.b*t1.a);

    return t;
}

int main()
{
    int N;

    while(scanf("%d", &N) != EOF)
    {
        segment p1, p2;
        point A, B;

        printf("INTERSECTING LINES OUTPUT\n");
        while(N--)
        {
            scanf("%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y);p1.InIt(A, B);
            scanf("%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y);p2.InIt(A, B);

            if(Collineation(p1, p2) == true)
                printf("LINE\n");
            else if(Parallel(p1, p2) == true)
                printf("NONE\n");
            else
            {
                point ans = CrossPoint(p1, p2);
                printf("POINT %.2f %.2f\n", ans.x, ans.y);
            }
        }
        printf("END OF OUTPUT\n");
    }

    return ;
}