//以下为原blog搬迁过来的内容

【题目大意】：给出一条线段和一个矩形࿰c;判断线段和矩形是否相交。

【解题思路】：判给出的线段与矩形的每条线段是否相交（包括相交和非规范相交）。

1、内含也算相交（WA一次）       2、看了discuss࿰c;发现题目给出矩形的点并不一定是左上角的点和右下角的点（WA2次）

【代码】：

<code class="language-cpp">#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <cctype>
#include <map>
#include <iomanip>
                   
using namespace std;
                   
#define eps 1e-8
#define pi acos(-1.0)
#define inf 1<<30
#define pb push_back
#define lc(x) (x << 1)
#define rc(x) (x << 1 | 1)
#define lowbit(x) (x & (-x))
#define ll long long

class="tags" href="/tags/STRUCT.html" title=struct>struct Point
{
    double x, y;
    Point() {}
    Point(double a, double b)
    {
        x = a, y = b;
    }
}point[20];

class="tags" href="/tags/STRUCT.html" title=struct>struct Line
{
    Point a, b;
    Line() {}
    Line(Point x, Point y)
    {
        a = x, b = y;
    }
}line,line1[20];

inline int sig(double k)
{
    return k < -eps ? -1 : k > eps;
}

inline double det(double x1, double y1, double x2, double y2)
{
    return x1 * y2 - x2 * y1;
}
inline double xmult(Point o, Point a, Point b)
{
    return det(a.x - o.x, a.y - o.y, b.x - o.x, b.y - o.y);

}

inline double dotdet(double x1, double y1, double x2, double y2)
{
    return x1 * x2 + y1 * y2;
}

inline double dot(Point o, Point a, Point b)
{
    return dotdet(a.x - o.x, a.y - o.y, b.x - o.x, b.y - o.y);
}

inline int between(Point o, Point a, Point b)
{
    return sig(dot(o, a, b));
}

inline int intersect1(Point a, Point b, Point c, Point d) {
    double s1, s2, s3, s4;
    int d1 = sig(s1 = xmult(a, b, c));
    int d2 = sig(s2 = xmult(a, b, d));
    int d3 = sig(s3 = xmult(c, d, a));
    int d4 = sig(s4 = xmult(c, d, b));
    if ((d1^d2) == -2 && (d3^d4) == -2)
    {
        return 1;
    }
    if (d1 == 0 && between(c, a, b) <= 0) return 2;
    if (d2 == 0 && between(d, a, b) <= 0) return 2;
    if (d3 == 0 && between(a, c, d) <= 0) return 2;
    if (d4 == 0 && between(b, c, d) <= 0) return 2;
    return 0;
}

inline int intersect(Line u, Line v)
{
    return intersect1(u.a, u.b, v.a, v.b);
}

int main()
{
    int T;
    double p,q,m,n,p1,q1,m1,n1;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%lf%lf",&p1,&q1);

        scanf("%lf%lf",&m1,&n1);
        point[1]=Point(m1,n1);
        point[0]=Point(p1,q1);
        line=Line(point[0],point[1]);
        scanf("%lf%lf%lf%lf",&p,&q,&m,&n);
        if (p>m && q<n)
        {
            swap(p,m);
            swap(q,n);
        }
        else
        if (p>m && q>n) swap(m,p);
        else
        if (p<m && q<n) swap(q,n);
        if ((p1>=p && p1<=m && q1<=q && q1>=n && m1>=p && m1<=m && n1<=q && n1>=n))
        {
            printf("T\n");
            continue;
        }
        point[2]=Point(p,q);
        point[3]=Point(m,q);
        point[4]=Point(p,n);
        point[5]=Point(m,n);
        line1[0]=Line(point[2],point[3]);
        line1[1]=Line(point[2],point[4]);
        line1[2]=Line(point[3],point[5]);
        line1[3]=Line(point[4],point[5]);
        int k1,k2,k3,k4;
        k1=intersect(line,line1[0]);
        k2=intersect(line,line1[1]);
        k3=intersect(line,line1[2]);
        k4=intersect(line,line1[3]);
        //cout << k1 << " " << k2 << " " << k3 << " " << k4 << endl;
        if (k1==0 && k2==0 && k3==0 && k4==0)
        {
            printf("F\n");
        }
        else printf("T\n");

    }
    return 0;
}
code>