POJ 1269 Intersecting Lines(直線位置關係，水)

利用叉積判斷直線位置關係，叉積3爲 0 則共線或平行，否則相交，設交點 O

則有 P1O × OP2 = P3O × OP4，可以推出公式。

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

const double eps = 1e-8;

typedef struct point {
    double x;
    double y;
}point;

typedef struct v {
    point start;
    point end;
}v;

double crossProduct(v* v1, v* v2)
{
    v vt1, vt2;
    double result = 0;

    vt1.start.x = 0;
    vt1.start.y = 0;
    vt1.end.x = v1->end.x - v1->start.x;
    vt1.end.y = v1->end.y - v1->start.y;

    vt2.start.x = 0;
    vt2.start.y = 0;
    vt2.end.x = v2->end.x - v2->start.x;
    vt2.end.y = v2->end.y - v2->start.y;

    result = vt1.end.x * vt2.end.y - vt1.end.y * vt2.end.x;
    return result;
}

void lineCross(point p1, point p2, point p3, point p4)
{
    v v1, v2, v3;
    v1.start = p1;
    v1.end = p2;
    v2.start = p3;
    v2.end = p4;
    v3.start = p2;
    v3.end = p3;
    if(fabs(crossProduct(&v1, &v2)) < eps) {
        if(fabs(crossProduct(&v2, &v3)) < eps)
            puts("LINE");   //same line
        else
            puts("NONE");   //parallel
    } else {
        double a1 = p1.y - p2.y, b1 = p2.x - p1.x, c1 = p1.x * p2.y - p2.x * p1.y;
        double a2 = p3.y - p4.y, b2 = p4.x - p3.x, c2 = p3.x * p4.y - p4.x * p3.y;
        double x = (b1*c2 - b2*c1) / (a1*b2 - a2*b1);
        double y = (a1*c2 - a2*c1) / (b1*a2 - b2*a1);
        printf("POINT %.2lf %.2lf\n", x, y);    //output the point
    }
}

int main()
{
    int n;

    puts("INTERSECTING LINES OUTPUT");
    for(scanf("%d", &n); n--;) {
        point p1, p2, p3, p4;
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y, &p4.x, &p4.y);
        lineCross(p1, p2, p3, p4);
    }
    puts("END OF OUTPUT");

    return 0;
}