題意：給出n條線段，問是否存在一條直線使所有線段在其上的對映有至少一個共點

假設找到了這條直線，那過共點作直線的垂線必然與n條線段相交，就相當於問是否存在直線可以與所有線段相交

\(n^2\)列舉直線，然後\(O(n)\)判斷

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e3+100;
const double eps=1e-8;
struct Point{
    double x,y;
    Point(double xx=0,double yy=0){
        x=xx,y=yy;
    }
}s[maxn],t[maxn];
struct Vector{
    double x,y;
    Vector(double xx=0,double yy=0){
        x=xx,y=yy;
    }
};
int dcmp(double x){return fabs(x)<eps?0:(x>0?1:-1);}
Vector operator - (Point a,Point b){return Vector(a.x-b.x,a.y-b.y);}
double operator * (Vector a,Vector b){return a.x*b.y-a.y*b.x;}
int tt,n;
/*bool segment(Point a,Point b,Point c,Point d){
    Vector x=b-a,y=d-c;
    Vector v1=c-a,v2=d-a;
    if(dcmp(x*v1)*dcmp(x*v2)>0) return 0;
    v1=a-c,v2=b-c;
    if(dcmp(y*v1)*dcmp(y*v2)>0) return 0;
    return 1;
}*/
bool check(Point x,Point y){
    if(dcmp(x.x-y.x)==0&&dcmp(x.y-y.y)==0) return 0;
    for(int i=1;i<=n;i++)
        if(((y-x)*(s[i]-x))*((y-x)*(t[i]-x))>eps)
            return 0;
    return 1;
}
int main(){
    scanf("%d",&tt);
    while(tt--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%lf%lf%lf%lf",&s[i].x,&s[i].y,&t[i].x,&t[i].y);
        bool ok=0;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++)
                if(check(s[i],t[j])||check(s[i],s[j])||check(t[i],t[j])||check(t[i],s[j])){
                    ok=1;
                    break;
                }
            if(ok) break;
        }
        if(ok) printf("Yes!\n");
        else printf("No!\n");
    }
    return 0;
}