题目描述
Two endpoints of two line segments on a plane are given to determine whether the two segments are intersected (there is a common point or there is a partial coincidence that intersects). If intersected, output "Yes", otherwise output "No".
输入描述:
The first line is a number of T, indicating the number of tests inputed (1 <= T <= 1000)
For next T line,each line contains 8 numbers , x1,y1,x2,y2,x3,y3,x4, y4. (8-10 ^ < = xi, yi < = 10 ^ 8)
(the two endpoints of line 1 are x1, y1, |, x2, y2, and two of the endpoints of line 2 are x3, y3, |, x4, y4).
输出描述:
For each test case, output"Yes" if the two segments intersected, else output"No".
示例1
输入
2 1 2 2 1 0 0 2 2 -1 1 1 1 0 0 1 -1
输出
Yes No
题解
判断线段非严格相交。
#include<cstdio>
using namespace std;
const double 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);
}
int dots_inline(point p1,point p2,point p3)
{
return zero(xmult(p1,p2,p3));
}
int same_side(point p1,point p2,point l1,point l2)
{
return xmult(l1,p1,l2)*xmult(l1,p2,l2)>eps;
}
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 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);
}
//调用intersect_in，相交返回1
int main() {
int T;
scanf("%d", &T);
while(T --) {
point a, b, c, d;
scanf("%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y);
scanf("%lf%lf%lf%lf", &c.x, &c.y, &d.x, &d.y);
if(intersect_in(a,b,c,d)) printf("Yes\n");
else printf("No\n");
}
return 0;
}