Algorithm

D. Pair Of Lines
time limit per test
2 seconds
memory limit per test
256 megabytes
Problem link- https://codeforces.com/contest/961/problem/D
input
standard input
output
standard output

You are given n points on Cartesian plane. Every point is a lattice point (i. e. both of its coordinates are integers), and all points are distinct.

You may draw two straight lines (not necessarily distinct). Is it possible to do this in such a way that every point lies on at least one of these lines?

Input

The first line contains one integer n (1 ≤ n ≤ 105) — the number of points you are given.

Then n lines follow, each line containing two integers xi and yi (|xi|, |yi| ≤ 109)— coordinates of i-th point. All n points are distinct.

Output

If it is possible to draw two straight lines in such a way that each of given points belongs to at least one of these lines, print YES. Otherwise, print NO.

Examples
input
Copy
5
0 0
0 1
1 1
1 -1
2 2
output
Copy
YES
input
Copy
5
0 0
1 0
2 1
1 1
2 3
output
Copy
NO
Note

In the first example it is possible to draw two lines, the one containing the points 1, 3 and 5, and another one containing two remaining points.

 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int const N = 1e5 + 1;
int n, a[N], b[N];
bool vis[N];

bool solve(int i1, int i2) {
	memset(vis, false, sizeof vis);

	long long s1, s2;
  int v1 = -1, v2 = -1;

  s1 = b[i2] - b[i1];
  s2 = a[i2] - a[i1];

  vis[i1] = vis[i2] = true;
  int cnt = 2;
  for(int i = 0; i < n; ++i) {
  	if(vis[i])
  		continue;

  	long long ss1, ss2;
  	ss1 = b[i] - b[i1];
  	ss2 = a[i] - a[i1];

  	if(s2 * ss1 == s1 * ss2) {
  		vis[i] = true;
  		++cnt;
  	} else {
  		if(v1 == -1)
  			v1 = i;
  		else if(v2 == -1)
  			v2 = i;
  	}
  }

  if(n - cnt <= 2) {
  	return true;
  }

  long long e1, e2;

  e1 = b[v2] - b[v1];
  e2 = a[v2] - a[v1];
  vis[v1] = vis[v2] = true;
  cnt += 2;

  for(int i = 0; i < n; ++i) {
  	if(vis[i])
  		continue;

  	long long ss1, ss2;
  	ss1 = b[i] - b[v1];
  	ss2 = a[i] - a[v1];

  	if(e2 * ss1 == e1 * ss2) {
  		vis[i] = true;
  		++cnt;
  	}
  }

  if(cnt == n) {
  	return true;
  }

  return false;
}

int main() {
  cin >> n;
  for(int i = 0; i < n; ++i)
  	cin >> a[i] >> b[i];

  if(n <= 2) {
  	cout << "YES" << endl;
  	return 0;
  }

  if(solve(0, 1) || solve(0, 2) || solve(1, 2))
  	cout << "YES" << endl;
  else
  	cout << "NO" << endl;

  return 0;
}
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
5
0 0
0 1
1 1
1 -1
2 2

Output

x
–
+
cmd
YES
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Demonstration

Codeforoces Solution-Pair Of Lines-Solution in C, C++, Java, Python,Codeforoces Solution

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