Segments

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 x₂y₂ 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<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
//#include<bits/stdc++.h>
#define eps 1e-8
using namespace std;
struct Line
{
   double x1,y1;
   double x2,y2;
}line[110];
double dis(double x1,double y1,double x2,double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double cross(double x1,double y1,double x2,double y2,double x,double y)
{
    return (x2-x1)*(y-y1)-(x-x1)*(y2-y1);
}
int t,n;
bool  jude(double x1,double y1,double x2,double y2)
{
    if(dis(x1,y1,x2,y2)<eps) return 0;
    for(int i=0;i<n;i++)
    {
        if(cross(x1,y1,x2,y2,line[i].x1,line[i].y1)*
           cross(x1,y1,x2,y2,line[i].x2,line[i].y2)>eps)
            return 0;
    }
    return 1;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&line[i].x1,&line[i].y1,&line[i].x2,&line[i].y2);
        }
        if(n<3)
        {
            printf("Yes!\n");
            continue;
        }
        int flag=0;
      for(int i=0;i<n&&!flag;i++)
      {
          for(int j=i+1;j<n&&!flag;j++)
          {
              if(jude(line[i].x1,line[i].y1,line[j].x1,line[j].y1)||
                 jude(line[i].x1,line[i].y1,line[j].x2,line[j].y2)||
                 jude(line[i].x2,line[i].y2,line[j].x1,line[j].y1)||
                 jude(line[i].x2,line[i].y2,line[j].x2,line[j].y2))
                    flag=1;
          }
      }
      if(flag)
        printf("Yes!\n");
      else
        printf("No!\n");


    }
}