HDU_1086 You can Solve a Geometry Problem too（几何题）

计算几何讲解：http://dev.gameres.com/Program/Abstract/Geometry.htm#判断两线段是否相交

一个比较好的范例：http://www.wutianqi.com/?p=2407

代码：

#include <stdio.h>
struct node
{
    double x1, y1;
    double x2, y2;
}l[110];

double X(double a, double b, double c, double d)
{
    return a*d - b*c;
}

int judge(int i, int j)
{
    double a = X(l[i].x1-l[j].x1, l[i].y1-l[j].y1, l[j].x2-l[j].x1, l[j].y2-l[j].y1);
    double b = X(l[i].x2-l[j].x1, l[i].y2-l[j].y1, l[j].x2-l[j].x1, l[j].y2-l[j].y1);
    double c = a*b;
    double d = X(l[j].x1-l[i].x1, l[j].y1-l[i].y1, l[i].x2-l[i].x1, l[i].y2-l[i].y1);
    double e = X(l[j].x2-l[i].x1, l[j].y2-l[i].y1, l[i].x2-l[i].x1, l[i].y2-l[i].y1);
    double f = d*e;
    if(c <= 0 && f <= 0)
        return 1;
    return 0;
}

int main()
{
    int n, i, j;
    //freopen("data.in", "r", stdin);
    while(scanf("%d", &n), n)
    {
        int ans = 0;
        for(i = 1; i <= n; i++)
            scanf("%lf%lf%lf%lf", &l[i].x1, &l[i].y1, &l[i].x2, &l[i].y2);
        for(i = 1; i < n; i++)
            for(j = i+1; j <= n; j++)
            {
                if(judge(i, j))
                    ans++;
            }
        printf("%d\n", ans);
    }
    return 0;
}