【題目】

傳送門

Description

Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common.

Input

Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing a positive integer n ≤ 100 showing the number of segments. After that, n lines containing four real numbers x1 y1 x2 y2 follow, in which (x1, y1) and (x2, y2) are the coordinates of the two endpoints for one of the segments.

Output

For each test case, your program must output “Yes!”, if a line with desired property exists and must output “No!” otherwise. You must assume that two floating point numbers a and b are equal if |a - b| < 10-8.

Sample Input

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

Sample Output

Yes!
Yes!
No!

【分析】

大致題意：給出 $n$ 條線段，問是否存在一條直線，使得所有的線段投影到它上面後，至少有一個公共交點

轉換一下題意，假設所有的存在這樣的直線，那麼在公共部分做一條垂線，必然能經過所有線段，所以原問題就可以轉換為是否存在一條直線，使它能經過所有線段

可以發現，如果存在這樣的直線，那麼它必然可以恰好經過兩條線段的端點（就是理解成這條線可以在允許的範圍內旋轉，而這兩個端點恰好卡住了這條線）

因此列舉這兩個端點，然後暴力列舉求解即可

【程式碼】

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 105
#define eps 1e-8
using namespace std;
int n;
struct point
{
	double x,y;
	point(){}
	point(double x,double y):x(x),y(y){}
	point operator+(const point &a)  {return point(x+a.x,y+a.y);}
	point operator-(const point &a)  {return point(x-a.x,y-a.y);}
	friend double dot(const point &a,const point &b)  {return a.x*b.x+a.y*b.y;}
	friend double cross(const point &a,const point &b)  {return a.x*b.y-b.x*a.y;}
	friend double dist(const point &a,const point &b)  {return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
};
struct line
{
	point s,t;
	line(){}
	line(point s,point t):s(s),t(t){}
}l[N];
int sign(double x)
{
	if(fabs(x)<eps)  return 0;
	if(x<0)  return -1;
	return 1;
}
bool intersect(line a,line b)
{
	return sign(cross(b.s-a.s,b.s-a.t)*cross(b.t-a.s,b.t-a.t))<=0;
}
bool check(line L)
{
	if(sign(dist(L.s,L.t))==0)
	  return false;
	for(int i=1;i<=n;++i)
	  if(!intersect(L,l[i]))
	    return false;
	return true;
}
int main()
{
	int i,j,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;++i)
		  scanf("%lf%lf%lf%lf",&l[i].s.x,&l[i].s.y,&l[i].t.x,&l[i].t.y);
		bool flag=false;
		if(n==1)  flag=true;
		for(i=1;i<=n;++i)
		  if(check(l[i]))
		    flag=true;
		for(i=1;i<=n;++i)
		{
			for(j=i+1;j<=n;++j)
			{
				if(check(line(l[i].s,l[j].s)))  flag=true;
				if(check(line(l[i].s,l[j].t)))  flag=true;
				if(check(line(l[i].t,l[j].s)))  flag=true;
				if(check(line(l[i].t,l[j].t)))  flag=true;
				if(flag)  break;
			}
			if(flag)  break;
		}
		puts(flag?"Yes!":"No!");
	}
	return 0;
}