ANTTT.cpp

//BISMILLAHIRRAHMANIRRAHIM
/*
 manus tar shopner soman boro
 Author :: Shakil Ahmed
.............AUST_CSE27.........
 prob   ::
 Type   ::
 verdict::
 */

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define pi acos(-1.0)
#define ff first
#define ss second
#define re return
#define QI queue<int>
#define SI stack<int>
#define SZ(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define sqr(x) ((x) * (x))
#define ms(a,b) memset((a),(b),sizeof(a))
#define G() getchar()
#define MAX3(a,b,c) max(a,max(b,c))
#define II ( { int a ; read(a) ; a; } )
#define LL ( { Long a ; read(a) ; a; } )
#define DD ({double a; scanf("%lf", &a); a;})

double const EPS=3e-8;
using namespace std;

#define FI freopen ("input.txt", "r", stdin)
#define FO freopen ("output.txt", "w", stdout)

typedef long long Long;
typedef long long int64;
typedef unsigned long long ull;
typedef vector<int> vi ;
typedef set<int> si;
typedef vector<Long>vl;
typedef pair<int,int>pii;
typedef pair<string,int>psi;
typedef pair<Long,Long>pll;
typedef pair<double,double>pdd;
typedef vector<pii> vpii;

// For loop

#define forab(i, a, b)	for (__typeof (b) i = (a) ; i <= b ; ++i)
#define rep(i, n)		forab (i, 0, (n) - 1)
#define For(i, n)		forab (i, 1, n)
#define rofba(i, a, b)	for (__typeof (b)i = (b) ; i >= a ; --i)
#define per(i, n)		rofba (i, 0, (n) - 1)
#define rof(i, n)		rofba (i, 1, n)
#define forstl(i, s)	for (__typeof ((s).end ()) i = (s).begin (); i != (s).end (); ++i)

template< class T > T gcd(T a, T b) { return (b != 0 ? gcd<T>(b, a%b) : a); }
template< class T > T lcm(T a, T b) { return (a / gcd<T>(a, b) * b); }

//Fast Reader
template<class T>inline bool read(T &x){int c=getchar();int sgn=1;while(~c&&c<'0'||c>'9'){if(c=='-')sgn=-1;c=getchar();}for(x=0;~c&&'0'<=c&&c<='9';c=getchar())x=x*10+c-'0'; x*=sgn; return ~c;}

//int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction
//int dx[]={1,1,0,-1,-1,-1,0,1};int dy[]={0,1,1,1,0,-1,-1,-1};//8 direction
//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction
//int dx[]={2,1,-1,-2,-1,1};int dy[]={0,1,1,0,-1,-1}; //Hexagonal Direction

/* **************************************  My code start here ****************************************** */
const int MX = 1005 ;
struct Point
{
    int x , y ;
};

struct abc
{
    Point s , e ;
}inp[MX];

bool onSegment(Point p, Point q, Point r)
{
    if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) &&
        q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y))
       return true;

    return false;
}

// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are colinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p, Point q, Point r)
{
    // See 10th slides from following link for derivation of the formula
    // http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf
    int val = (q.y - p.y) * (r.x - q.x) -
              (q.x - p.x) * (r.y - q.y);

    if (val == 0) return 0;  // colinear

    return (val > 0)? 1: 2; // clock or counterclock wise
}

// The main function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
bool doIntersect(Point p1, Point q1, Point p2, Point q2)
{
    // Find the four orientations needed for general and
    // special cases
    int o1 = orientation(p1, q1, p2);
    int o2 = orientation(p1, q1, q2);
    int o3 = orientation(p2, q2, p1);
    int o4 = orientation(p2, q2, q1);

    // General case
    if (o1 != o2 && o3 != o4)
        return true;

    // Special Cases
    // p1, q1 and p2 are colinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(p1, p2, q1)) return true;

    // p1, q1 and p2 are colinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(p1, q2, q1)) return true;

    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(p2, p1, q2)) return true;

     // p2, q2 and q1 are colinear and q1 lies on segment p2q2
    if (o4 == 0 && onSegment(p2, q1, q2)) return true;

    return false; // Doesn't fall in any of the above cases
}



int n , m , color[MX] , adj[MX][MX] , deg[MX] ;
void dfs( int x , int c)
{
    color[x] = c ;
    rep( i , deg[x] )
    {
        int v = adj[x][i];
        if( color[v] == 0 ) dfs( v , c ) ;
    }
}
void solve()
{
    int c = 0 ;
    rep ( i , n ) if( color[i] == 0 ) dfs( i , ++c );
    rep( i , m )
    {
        int x = II , y = II ;
        puts(color[x-1] == color[y-1] ? "YES" : "NO" );
    }
}
void input()
{
    n = II , m = II ;
    rep( i , n )
    {
        inp[i].s.x = II , inp[i].s.y = II ;
        inp[i].e.x = II , inp[i].e.y = II ;
        color[i] = 0 ;
        deg[i] = 0 ;

    }
}
void Graph()
{
    int i , j ;
    for ( i = 0 ; i < n ; i++ )
    {
        for ( j = i + 1 ; j < n ; j++ )
        {
            if( doIntersect(inp[i].s , inp[i].e , inp[j].s , inp[j].e ) )
            {
                adj[i][deg[i]++] = j ;
                adj[j][deg[j]++] = i ;
            }
        }
    }
}
int main()
{
   // I will always use scanf and printf
   // May be i won't be a good programmer but i will be a good human being
     int cs , t = II ;
     for ( cs = 1 ; cs <= t ; cs++ )
     {

               input();
               Graph();
               solve();
     }
    return 0;
}