3
2
1.0 2.0 3.0 4.0
4.0 5.0 6.0 7.0
3
0.0 0.0 0.0 1.0
0.0 1.0 0.0 2.0
1.0 1.0 2.0 1.0
3
0.0 0.0 0.0 1.0
0.0 2.0 0.0 3.0
1.0 1.0 2.0 1.0

Yes!
Yes!
No!

显然该题可转换为是否存在一条直线与所有线段相交。（如果存在，则通过移动选转，使其过的2个线段的端点）

需要特判2个点如果相同，那么连不成线段，因为最后叉积算出来为0.

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;
#define eps 1e-8
#define MAXN (200+10)
struct P
{
	long double x,y;
	P(){}
	P(long double _x,long double _y):x(_x),y(_y){}
}a[MAXN*2];
struct V
{
	long double x,y;
	V(){}
	V(long double _x,long double _y):x(_x),y(_y){}
	V(P a,P b):x(b.x-a.x),y(b.y-a.y){}
};
struct S
{
	P A,B;
	S(P _A,P _B):A(_A),B(_B){}
	S(){}
};
double operator*(V a,V b)
{
	return a.x*b.y-b.x*a.y;
}
bool corr(P A,P B,P C,P D)
{
	V AB=V(A,B),AC=V(A,C),AD=V(A,D);
	return ((AC*AB)*(AB*AD)>-eps);
}
bool fabs (double a,double b)
{
	if (abs(a-b)<eps) return 1 ;return 0;
}
bool fabs(P A,P B)
{
	return (fabs(A.x,B.x)&&fabs(A.y,B.y));
}
int t,n;
int main()
{
//	freopen("poj3304.in","r",stdin);
	scanf("%d",&t);
	while (t--)
	{
		cin>>n;
		bool flag=0;
		for (int i=1;i<=2*n;i++) cin>>a[i].x>>a[i].y;
		for (int i=1;i<2*n&&!flag;i++)
		{
			for (int j=i+1;j<=2*n;j++)
			{
				if (fabs(a[i],a[j])) continue;
				S l=S(a[i],a[j]);
				int k;
				for (k=1;k<=n;k++)
				{
					if (!corr(l.A,l.B,a[k*2-1],a[k*2])) break;
				//	cout<<k;
				}
				if (k==n+1) {cout<<"Yes!n"; flag=1;break;}
			}
		}

		if (!flag) cout<<"No!n";


	}
	return 0;
}

5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5

INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT

模板如下：

注意* 表示叉积

这题涉及已知相交，线段跨立求交点

异侧情况：

同侧情况：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define eps 1e-8
double sqr(double x) {return x*x;}
struct P
{
	double x,y;
	P(double _x,double _y):x(_x),y(_y){}
	P(){}
	double dis()
	{
		return sqrt(sqr(x)+sqr(y));
	}
};
struct V
{
	double x,y;
	V(double _x,double _y):x(_x),y(_y){}
	V(P a,P b):x(b.x-a.x),y(b.y-a.y){}
	V(){}
	const double dis()
	{
		return sqrt(sqr(x)+sqr(y));
	}
};
P operator+(const P a,const V b)
{
	return P(a.x+b.x,a.y+b.y);
}
V operator*(const double a,const V b)
{
	return V(a*b.x,a*b.y);
}
double operator*(const V a,const V b)
{
	return a.x*b.y-b.x*a.y;
}
P jiao_dian(const V a,V b,const V c,const V CD,const P C)
{
	double d;
	d=b.dis();
	double s1=a*b,s2=b*c;
	double k=s1/(s1+s2);
	return C+k*CD;
}
bool equal(const double a,const double b)
{
	if (abs(a-b)<eps) return 1;return 0;
}
int n;
int main()
{
//s	freopen("poj1269.in","r",stdin);
	cout<<"INTERSECTING LINES OUTPUT"<<endl;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		double x1,y1,x2,y2,x3,y3,x4,y4;
		scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
		P A=P(x1,y1),B=P(x2,y2),C=P(x3,y3),D=P(x4,y4);
		V AB=V(A,B),AC=V(A,C),AD=V(A,D),CD=V(C,D);
		if (equal((AB*CD),0))
		{
			if (equal((AC*AD),0)) cout<<"LINEn";
			else cout<<"NONEn";
		}
		else
		{
			P p=jiao_dian(AC,AB,AD,CD,C);
			cout.setf(ios::fixed);
			cout.precision(2);
			cout<<"POINT "<<p.x<<' '<<p.y<<endl;
		}
	}
	cout<<"END OF OUTPUT"<<endl;
	return 0;
}