给你一个矩形的左下点和右上点的坐标,以及一条线段的两个端点的坐标,判断线段是否与矩形相交。相交输出T,否则输出F。
输入:
1
4 9 11 2 1 5 7 1
输出:
F
思路:
对矩形的四条边分别和线段进行叉积判断是否相交
代码:
#include"math.h"
#include"stdio.h"
#include"string.h"
#include"algorithm"
using namespace std;
struct M
{
double x,y;
}a1,a2,b1,b2,b3,b4;
double mul(M a,M b,M c)
{
return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
bool in(M a,M b,M c,M d)
{
if(max(a.x,b.x)<min(c.x,d.x)||max(c.x,d.x)<min(a.x,b.x))
return 0;
if(max(a.y,b.y)<min(c.y,d.y)||max(c.y,d.y)<min(a.y,b.y))
return 0;
if(mul(a,d,c)*mul(d,b,c)<0)
return 0;
if(mul(c,b,a)*mul(b,d,a)<0)
return 0;
return 1;
}
bool judege(M a,M b,M c,M d,M e,M f)
{
if(in(e,f,a,b))
return 1;
if(in(e,f,b,c))
return 1;
if(in(e,f,c,d))
return 1;
if(in(e,f,d,a))
return 1;
return 0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int flag=0;
scanf("%lf %lf %lf %lf %lf %lf %lf %lf",&a1.x,&a1.y,&a2.x,&a2.y,&b1.x,&b1.y,&b3.x,&b3.y);
if(b1.x>b3.x)
swap(b1.x,b3.x);
if(b1.y<b3.y)
swap(b1.y,b3.y);
b2.x=b1.x;
b2.y=b3.y;
b4.x=b3.x;
b4.y=b1.y;
if(judege(b1,b2,b3,b4,a1,a2))
flag=1;
if((b1.x<=a1.x&&a1.x<=b3.x&&b3.y<=a1.y&&a1.y<=b1.y)||(b1.x<=a2.x&&a2.x<=b3.x&&b3.y<=a2.y&&a2.y<=b1.y))
flag=1;
if(flag)
printf("T\n");
else
printf("F\n");
}
return 0;
}