题目大意：
题目意思很简单，就是说有一个矩阵是实心的，给出一条线段，问线段和矩阵是否相交
An example:
line: start point: (4,9)
end point: (11,2)
rectangle: left-top: (1,5)
right-bottom: (7,1) 
Sample Input
1
4 9 11 2 1 5 7 1
Sample Output
F
题解：
一道鬼题
本身不难，只要拿四条边与线段做快速排斥和跨立就行
但有很多细节：
1.给出的矩形的坐标要判断，不合法要交换
2.包含在矩形内，不与任何边相交
3.与边共线，还要判断是否有重合部分
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<iostream> 5 #include<cmath> 6 using namespace std; 7 struct Node 8 { 9 double x1,y1,x2,y2; 10 }line[10001]; 11 int T; 12 double xe,ye,xs,ys,xl,yt,xr,yb; 13 double direction(double x,double y,double x1,double y1,double x2,double y2) 14 { 15 double a1=x1-x; 16 double b1=y1-y; 17 double a2=x2-x; 18 double b2=y2-y; 19 return a1*b2-a2*b1; 20 } 21 int on_segment(double x1,double y1,double x2,double y2,double x,double y) 22 { 23 if((min(x1,x2)<=x&&x<=max(x1,x2))&&(min(y1,y2)<=y&&y<=max(y1,y2))) 24 return 1; 25 return 0; 26 } 27 28 bool exam(Node v,Node t) 29 { 30 double d1,d2,d3,d4; 31 d1=direction(t.x1,t.y1,t.x2,t.y2,v.x1,v.y1); 32 d2=direction(t.x1,t.y1,t.x2,t.y2,v.x2,v.y2); 33 d3=direction(v.x1,v.y1,v.x2,v.y2,t.x1,t.y1); 34 d4=direction(v.x1,v.y1,v.x2,v.y2,t.x2,t.y2); 35 if(d1*d2<0 && d3*d4<0) return 1; 36 if(!d1&&on_segment(t.x1,t.y1,t.x2,t.y2,v.x1,v.y1)) return 1; 37 if(!d2&&on_segment(t.x1,t.y1,t.x2,t.y2,v.x2,v.y2)) return 1; 38 if(!d3&&on_segment(v.x1,v.y1,v.x2,v.y2,t.x1,t.y1)) return 1; 39 if(!d4&&on_segment(v.x1,v.y1,v.x2,v.y2,t.x2,t.y2)) return 1; 40 return 0; 41 } 42 int main() 43 { 44 cin>>T; 45 while (T--) 46 { 47 scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&xs,&ys,&xe,&ye,&xl,&yt,&xr,&yb); 48 if (yt<yb) swap(yt,yb); 49 if (xl>xr) swap(xl,xr); 50 Node d; 51 d.x1=xs;d.y1=ys; 52 d.x2=xe;d.y2=ye; 53 line[1].x1=xl;line[1].y1=yt;line[1].x2=xr;line[1].y2=yt; 54 line[2].x1=xl;line[2].y1=yb;line[2].x2=xr;line[2].y2=yb; 55 line[3].x1=xl;line[3].y1=yt;line[3].x2=xl;line[3].y2=yb; 56 line[4].x1=xr;line[4].y1=yt;line[4].x2=xr;line[4].y2=yb; 57 if (exam(d,line[1])||exam(d,line[2])||exam(d,line[3])||exam(d,line[4])) 58 { 59 cout<<"T "; 60 } 61 else 62 { 63 if (xl<=min(xs,xe)&&xr>=max(xs,xe)&&yt>=max(ys,ye)&&yb<=min(ys,ye)) 64 cout<<"T "; 65 else 66 cout<<"F "; 67 } 68 } 69 }