Getting run error in all the test cases

#1

#include
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main () {
ll ans=INT_MIN;
ll n,m;
ll dp[n+1][n+1];
ll x,y;

cin>>n>>m;
for(ll i=0;i<m;i++){
   cin>>x>>y;
   dp[x-1][y-1]++;
}
for(int i=1;i<n;i++){
	for(int j=1;j<n;j++){
		dp[i][j]+=dp[i-1][j];
	}
}
for(int i=1;i<n;i++){
	for(int j=1;j<n;j++){
		dp[i][j]+=dp[i][j-1];
	}
}
for(int i=1;i<n;i++){
	for(int j=1;j<n;j++){
       ans=max(ans,min(dp[i][j],min(dp[n-1][j]-dp[i][j],min(dp[i][n-1]-dp[i][j],dp[n-1][n-1]-dp[n-1][j]-dp[i][n-1]+dp[i][j]))));			
	}

}
cout<<ans<<endl;
return 0;
}
Getting run error in all the test cases.

#2

https://ide.codingblocks.com/s/329215 here is the link of the code.

#3

Just take prefix sum of every row and after that take prefix sum of every column.
Then if you partition the (i,j)th of grid then your top-left box sum will be the value at (i,j)th index of grid
and top-right box sum will be the sum at that row’s last column - the sum of top-left box sum
and bottom right box sum will be the value at (n,j)th of the grid - the sum of top-left box
and bottom-right box sum will be the value at (n,m)th of grid - top-right box sum - bottom left box sum + the sum of the top-left box ( because it gets subtracted two times, the first time in top-right box sum and the second time in bottom-left box sum ).
Make ans at every step as
ans = max ( ans, min( all four boxes) );
You can see my implementation, I have commented on the part for better understanding. It’s better if you try on a notebook with one small example to have a crystal clear understanding.