Segments

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 x₁y₁x₂y₂ 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!

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

#define MAX(a,b) ((a)>(b) ? (a):(b))
#define MIN(a,b) ((a)<(b) ? (a):(b))

int n;

struct point{
	double x,y;
};

struct line{
	point a,b;
}lin[110];

double judge(point a,point b,point c)
{
	return (b.x-a.x)*(c.y-b.y)-(c.x-b.x)*(b.y-a.y);
}

double dis(point a,point b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

bool intersect(double x,double y,double x1,double y1)
{
	point a,b;
	a.x=x; a.y=y; b.x=x1; b.y=y1;
	if(dis(a,b)<1e-8) return 0;
	for(int i=0;i<n;i++)
	{
		if(judge(a,b,lin[i].a)*judge(a,b,lin[i].b)>0) return 0;
	}
	return 1;
}

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int i,j,flag=1;
		scanf("%d",&n);
		for(i=0;i<n;i++) scanf("%lf%lf%lf%lf",&lin[i].a.x,&lin[i].a.y,&lin[i].b.x,&lin[i].b.y);
		if(n==1) {printf("Yes!\n"); continue;}
		for(i=0;i<n&&flag;i++)
		 for(j=i+1;j<n&&flag;j++)
		 {
			if(intersect(lin[i].a.x,lin[i].a.y,lin[j].a.x,lin[j].a.y)||
			   intersect(lin[i].a.x,lin[i].a.y,lin[j].b.x,lin[j].b.y)||
			   intersect(lin[i].b.x,lin[i].b.y,lin[j].a.x,lin[j].a.y)||
			   intersect(lin[i].b.x,lin[i].b.y,lin[j].b.x,lin[j].b.y))
			{
				printf("Yes!\n");
				flag=0;
			}
		 }
		if(flag) printf("No!\n");
	}
	return 0;
}