链接：http://poj.org/problem?id=1269

给你两条直线，判断相交不相交．相交求交点．
都用叉积来解决就好了．
先用叉积来判断两条直线是否平行，平行的话要么是重合，要么是不重合，再取任意一点用叉积判断这点是否在另一条直线上，在的话就是重合了．

不平行的话，就会有交点，假设交点为P_0(x_0,y_0)．
由叉积原理，有

\\begin{cases}
(P_1-P_0)\\times(P_2-P_0) = 0\\\\
(P_3-P_0)\\times(P_2-P_0) = 0\\\\
\\end{cases}

展开得

\\begin{cases}
(y_1-y_2)x_0+(x_2-x_1)y_0+x_1y_2-x_2y_1=0\\\\
(y_3-y_4)x_0+(x_4-x_3)y_0+x_3y_4-x_4y_3=0\\\\
\\end{cases}

解出 x_0,y_0即可

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const double EPS = 1e-9;
const int N = 10;
int sgn(double x)
{
    if (fabs(x) < EPS)
        return 0;
    return x < 0 ? -1 : 1;
}
struct Point {
    double x, y;
    Point(){}
    Point(double x, double y): x(x), y(y) {}
    void read() {
        scanf("%lf %lf", &x, &y);
    }
    Point operator + (const Point &a) const {
        return Point(x + a.x, y + a.y);
    }
    Point operator - (const Point &a) const {
        return Point(x - a.x, y - a.y);
    }
} P[N];
double cross(Point a, Point b)
{
    return a.x * b.y - a.y * b.x;
}
struct Line {
    Point s, e;
    Line(){}
    Line(Point a, Point b) {
        s = a, e = b;
    }
    pair<Point, int> operator & (const Line &b) const {
        Point res = s;
        if (sgn(cross(s - e, b.s - b.e)) == 0) {
            return sgn(cross(b.s - s, b.e - s)) ? make_pair(res, 1) : make_pair(res, 0);
        }
        double t = 1.0 * cross(s - b.s, b.s - b.e) / cross(s - e, b.s - b.e);
        res.x += (e.x - s.x) * t;
        res.y += (e.y - s.y) * t;
        return make_pair(res, 2);
    }
} line[2];
int main()
{
    printf("INTERSECTING LINES OUTPUT\n");
    int n;
    scanf("%d", &n);
    while (n--) {
        for (int i = 1; i <= 4; i++) {
            P[i].read();
        }
        line[0] = Line(P[1], P[2]);
        line[1] = Line(P[3], P[4]);
        pair<Point, int> ans = line[0] & line[1];
        if (ans.second == 0) {
            printf("LINE\n");
        } else if (ans.second == 1) {
            printf("NONE\n");
        } else {
            printf("POINT %.2lf %.2lf\n", ans.first.x, ans.first.y);
        }
    }
    printf("END OF OUTPUT\n");
    return 0;
}