Lighting Rectangle CodeChef SnackDown 2021 Advance Practice Contest

Question

You are given an axis-aligned rectangle in a 2D Cartesian plane. The bottom left corner of this rectangle has coordinates $(0, 0)$ and the top right corner has coordinates $(N - 1, N - 1)$ . You are also given $K$ light sources; each light source is a point inside or on the perimeter of the rectangle.

For each light source, let's divide the plane into four quadrants by a horizontal and a vertical line passing through this light source. The light source can only illuminate one of these quadrants (including its border, i.e. the point containing the light source and two half-lines), but the quadrants illuminated by different light sources may be different.

You want to assign a quadrant to each light source in such a way that when they illuminate their respective quadrants, the entire rectangle (including its perimeter) is illuminated. Find out whether it is possible to assign quadrants to light sources in such a way.

Input

The first line of the input contains an integer $T$ denoting the number of test cases. The description of the test cases follows.
The first line of each test case contains two space-separated integers $K$ and $N$ .
Each of the next $K$ lines contains two space-separated integers $x$ and $y$ denoting a light source with coordinates $(x, y)$ .

Output

For each test case, print a single line containing the string "yes" if it is possible to illuminate the whole rectangle or "no" if it is impossible.

Constraints

$1 \leq T \leq 5, 000$
$1 \leq K \leq 100$
$1 \leq N \leq 10^{9}$
$0 \leq x, y \leq N - 1$
no two light sources coincide

Example Input

2 10

0 0

1 0

2 10

1 2

1 1

Example Output

yes

Program Code in C++

#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

using namespace __gnu_pbds;

#define ll long long int

#define pb push_back

#define mp make_pair

#define all(x) x.begin(),x.end()

#define Max 100000000000000

#define min_heap priority_queue <ll, vector<ll>, greater<ll> >

template <typename T>

using ordered_set = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

int main()

{

ll t;

cin>>t;

while(t--){

ll k,n;

cin>>k>>n;

ll x[k],y[k];

for(ll i=0;i<k;i++) cin>>x[i]>>y[i];

if(k>=4){

cout<<"yes"<<endl;

continue;

}

ll p=0,q=0,chk=0;

ll u[]={0,0,n-1,n-1};

ll v[]={0,n-1,0,n-1};

for(ll i=0;i<k;i++){

for(ll j=0;j<4;j++){

if(x[i]==u[j] && y[i]==v[j]) chk=1;

}

if(x[i]==0 || x[i]==n-1) p++;

if(y[i]==0 || y[i]==n-1) q++;

}

if(chk) cout<<"yes"<<endl;

else if(p>=2 || q>=2) cout<<"yes"<<endl;

else if(p+q>=2 && k==3) cout<<"yes"<<endl;

else if(p+q==0 || k<3) cout<<"no"<<endl;

else{

pair<ll,ll> c[3];

for(ll i=0;i<3;i++) c[i]={x[i],y[i]};

if(p) for(ll i=0;i<3;i++) swap(c[i].first,c[i].second);

sort(c,c+3);

if(c[0].second==0 || c[0].second==n-1 || c[2].second==0 || c[2].second==n-1 || c[0].first==c[1].first || c[1].first==c[2].first) cout<<"yes"<<endl;

else cout<<"no"<<endl;

}

return 0;

}

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