Segment

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

# define EPS 1e-8

int n;

struct Segment
{
    double x1,x2,y1,y2;
}segment[1000];

double distance(double x1,double y1,double x2,double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

double fun(double x1,double y1,double x2,double y2,double x0,double y0)
{
    return  (x2-x1)*(y0-y1)-(x0-x1)*(y2-y1);
}

bool search(double x1,double y1,double x2,double y2)
{
    int i;
    //推断是不是同一条直线;
    if(distance(x1,y1,x2,y2)<EPS)    return false;
    //利用叉积定理。推断是不是和其它的线段相交：
    for(i=0;i<n;i++)
    {
        if(fun(x1,y1,x2,y2,segment[i].x1,segment[i].y1)*
           fun(x1,y1,x2,y2,segment[i].x2,segment[i].y2)>EPS)    return false;
    }
    return true;
}

int main ()
{
    //freopen("a.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int i;
        for(i=0;i<n;i++)
            scanf("%lf%lf%lf%lf",&segment[i].x1,&segment[i].y1,&segment[i].x2,&segment[i].y2);
        if(n==1)    { printf("Yes!
"); continue;}

        int ans=0;
        
        //将全部的线段都列举出来。
        for(i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
        {
            if(search(segment[i].x1,segment[i].y1,segment[j].x1,segment[j].y1)||
               search(segment[i].x1,segment[i].y1,segment[j].x2,segment[j].y2)||
               search(segment[i].x2,segment[i].y2,segment[j].x1,segment[j].y1)||
               search(segment[i].x2,segment[i].y2,segment[j].x2,segment[j].y2))
                ans=1;
        }
        if(ans)    printf("Yes!
");
        else    printf("No!
");
    }
    return 0;
}