My code works for small amounts of data, but not with big amounts of it?

Question

I want to say that most of the Code is well and functioning, BUT when there are bigger data, it just malfunctions. I got 30 point out of 100 for it so far.

Here are test inputs and outputs, the source code, and the exercise itself (roughly translated to English).

The row's first numbers are the rows of after the first line, and the second one is the number of datas that are in those rows. After that those are the datas that I need to work with. I store them in bemenet 2 dimensional array.

My exercise was to find out which rows had at least 2 non 0 numbers in it. And then report it in a Matrix which size is: the second number from the first row^2.

So because of this I searched which rows have at least 2 non 0 numbers in them, then I put them in a Vector, and what I tried to do was to have like a relation thing, when like for example 1,2, and 4 are selected, the output will be all 0s except the 1st row 2nd 1st row 4th, 2nd row 1st, 2nd row 4th, and 4th row 1st and 2nd. In the in1.txt it works just like this, and is perfect, however, for the in2. Every value except for the main diagonal will be 1. I don't know what I did wrong, and I've been trying for the last 2 hours to solve this.

Can someone try to help, or explain me what my error is? Any help would be appreciated! Here's my source code so far:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

struct adatok{
  int szam = 0;
  bool alkalmi = false;
};

bool IsZero(int a){
    if(a == 0) return true;
    else return false;
}

int indexOf(vector<auto> v, auto item){
  for(int i=0; i <v.size();++i)
     if(item == v[i]) return i;
  return -1;
}
bool contains(vector<auto> v, auto item){
    return indexOf(v,item) != -1;
}

int main()
{
    int n = 0, m = 0;
    cin >> n >> m;vector<int> vektor;
    adatok bemenet[n][m];
    for (int i = 0; i < n; i++){
        int seged = 0;
        for (int j = 0; j < m; j++){
            cin >> bemenet[i][j].szam;
            if (!IsZero(bemenet[i][j].szam)){
                seged++;
            }
        }
        if (seged >= 2){
            for(int j = 0; j < m; j++){
                if(!IsZero(bemenet[i][j].szam)){
                    vektor.push_back(j);
                }
            }
        }
    }

    sort( vektor.begin(), vektor.end() );
    vektor.erase( unique( vektor.begin(), vektor.end() ), vektor.end() );

    for (int i = 0; i < m; i++){
        for (int j = 0; j < m; j++){
            if (i == j) cout << "0";
            else if(contains(vektor, i) && contains(vektor, j)){
                cout << "1";
            }
            else cout << "0";
        }
        cout << "\n";
    }

    return 0;
}

Answer 1

It's the Variable Length Array adatok bemenet[n][m];. A large enough m and n, could be as low as a few hundred each, will exhaust the program's automatic storage and send the program off into the wild and wacky world of Undefined Behaviour.

Note that Variable Length Arrays are not a part of Standard C++ (See Why aren't variable-length arrays part of the C++ standard? for a discussion on why not) and are only added by extension on a few compilers. Use them with caution if at all.

Use something like the vector-based matrix structure proposed in this answer instead.

Answer 2

After deciding to start over fresh, and to turn on my brain, I came up with the solution. For my problem this was how I managed to solve it completely, using Cartesian product.

#include <iostream>
#include <array>
#include <string.h>
#include <vector>

using namespace std;

bool IsZero(int a){
    if(a == 0) return true;
    else return false;
}

int main()
{
    int n = 0, m = 0;
    cin >> n >> m;
    int bemenet[n][m];
    int kimenet[m][m];
    memset(kimenet, 0, m * m * sizeof(int));

    for (int i = 0; i < n; i++){
        vector<int> seged;
        int s1 = 0;
        for (int j = 0; j < m; j++){
            cin >> bemenet[i][j];
            if (!IsZero(bemenet[i][j])){
                seged.push_back(j);
            }
        }
        for (int j = 0; j < seged.size(); j++){
            for (int k = 0; k < seged.size(); k++){
                if (seged[j] != seged[k]){
                    kimenet[seged[j]][seged[k]] = 1;
                    kimenet[seged[k]][seged[j]] = 1;
                }
            }
        }
    }

    for (int i = 0; i < m; i++){
        for (int j = 0; j < m; j++){
            cout << kimenet[i][j];
        }
        cout << "\n";
    }

    return 0;
}