Output

簡單說就是先判斷線段是不是在矩形內，再分別判斷和四條邊是否相交...

判斷相交貌似一直是比較棘手的問題啊，不過這種線線相交還好...

#include<cstdio>
#include<cstring>
#include<cmath>
#define eps 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)
struct point
{
    double x, y;
};

double xmult(point p1, point p2, point p0)
{
    return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
}

double dmult(point p1, point p2, point p0)
{
    return (p1.x - p0.x)*(p2.x - p0.x) + (p1.y - p0.y)*(p2.y - p0.y);
}

int dots_inline(point p1, point p2, point p3)
{
    return zero(xmult(p1, p2, p3));
}

int dot_online_in(point p, point l1, point l2)
{
    return zero(xmult(p, l1, l2)) && (l1.x - p.x)*(l2.x - p.x)<eps && (l1.y - p.y)*(l2.y - p.y)<eps;
}

int same_side(point p1, point p2, point l1, point l2)
{
    return xmult(l1, p1, l2)*xmult(l1, p2, l2)>eps;
}

int intersect_in(point u1, point u2, point v1, point v2)
{
    if (!dots_inline(u1, u2, v1) || !dots_inline(u1, u2, v2))
        return !same_side(u1, u2, v1, v2) && !same_side(v1, v2, u1, u2);
    return dot_online_in(u1, v1, v2) || dot_online_in(u2, v1, v2) || dot_online_in(v1, u1, u2) || dot_online_in(v2, u1, u2);
}
int main()
{
    point x, y;
    point d1, d2, d3, d4;
    int cas, i, j, k;
    scanf("%d", &cas);
    while (cas--)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &x.x, &x.y, &y.x, &y.y, &d4.x, &d4.y, &d2.x, &d2.y);
        d1.x = d4.x, d1.y = d2.y;
        d3.x = d2.x, d3.y = d4.y;
        if ((x.x >= d2.x&&x.x <= d4.x) || (x.x >= d4.x&&x.x <= d2.x))
            if ((x.y >= d2.y&&x.y <= d4.y) || (x.y >= d4.y&&x.y <= d2.y))
            {
                printf("T\n");
                continue;
            }
        if ((y.x >= d2.x&&y.x <= d4.x) || (y.x >= d4.x&&y.x <= d2.x))
            if ((y.y >= d2.y&&y.y <= d4.y) || (y.y >= d4.y&&y.y <= d2.y))
            {
                printf("T\n");
                continue;
            }
        if (intersect_in(x, y, d1, d2) || intersect_in(x, y, d3, d2) || intersect_in(x, y, d3, d4) || intersect_in(x, y, d1, d4))
            printf("T\n");
        else printf("F\n");
    }
    return 0;
}