题目大意：给一个线段和一个矩形，判断线段是否和矩形有公共点。

 

分析：用矩形的四个边当线段判断与所给的线段是否有交点，需要注意的是给的矩形是不标准的，需要自己转换，还需要注意线段有可能在矩形内部。

 

代码如下：

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#include
      
       #include
       
        #include
        
         using namespace std;const int MAXN = 1e3+7;const int oo = 1e9+7;const double EPS = 1e-12;struct point{    double x, y;    point(double x=0, double y=0):x(x), y(y){}    point operator - (const point &t) const{        return point(x-t.x, y-t.y);    }    int operator * (const point &t) const{        double ans = x*t.y - y*t.x;        if(ans > EPS)return 1;        if(fabs(ans) < EPS)return 0;        return -1;    }};struct segment{    point A, B;    segment(point A=0, point B=0):A(A), B(B){}    bool Intersect(const segment &t)const{        int t1 = (A-B) * (t.A-B);        int t2 = (A-B) * (t.B-B);        if(t1==0 && t.A.x>=min(A.x, B.x) && t.A.x <= max(A.x, B.x)           && t.A.y>=min(A.y, B.y) && t.A.y <= max(A.y, B.y))return true;        if(t2==0 && t.B.x>=min(A.x, B.x) && t.B.x <= max(A.x, B.x)           && t.B.y>=min(A.y, B.y) && t.B.y <= max(A.y, B.y))return true;        if(t1==0&&t2==0 && max(t.A.x,t.B.x)>=max(A.x,B.x) &&min(t.A.x,t.B.x)<=min(A.x,B.x)           && max(t.A.y,t.B.y)>=max(A.y,B.y) &&min(t.A.y,t.B.y)<=min(A.y,B.y) )return true;        if(t1*t2 == -1)return true;        return false;    }};bool Find(segment a, segment sg[], int N){    for(int i=0; i
         
           B.x)swap(A.x, B.x);        if(A.y < B.y)swap(A.y, B.y);        sg[0] = segment(A, point(A.x, B.y));        sg[1] = segment(A, point(B.x, A.y));        sg[2] = segment(B, point(A.x, B.y));        sg[3] = segment(B, point(B.x, A.y));       /// printf("Case %d: ", t++);        if(Find(a, sg, 4) == true || (a.A.x>=A.x&&a.A.x<=B.x && a.A.y>=B.y && a.A.y <= A.y) )            printf("T\n");        else            printf("F\n");    }    return 0;}/**24 2 4 0 4 3 9 6**/