SPOJ PPATH solution with explanation

#include <bits/stdc++.h>
using namespace std;
const int N = 9999;
bool chk[N];
vector<int> graph[N];
int vis[N], level[N];
void sieve() // Generating the primes 
{
    int i, j;
    for (i = 2; i <= N; i++)
        chk[i] = true;

    for (i = 2; i * i <= N; i++)
    {
        if (chk[i] == true)
        {
            for (j = i * i; j <= N; j += i)
            {
                chk[j] = false;
            }
        }
    }
}
bool ok(int a, int b)  // check weather the primes has difference of one digit or not
{
    int cnt = 0;
    while (a)
    {

        if (a % 10 != b % 10)
            cnt++;
        a /= 10;
        b /= 10;
        if (cnt > 1)
            return false;
    }
    return true;
}
int bfs(int source, int destination)  //applying the bfs 
{
    queue<int> q;
    q.push(source);
    vis[source] = 1;
    while (!q.empty())
    {
        int cur = q.front();
        q.pop();
        for (auto child : graph[cur])
        {
            if (!vis[child])
            {
                q.push(child);
                vis[child] = 1;
                level[child] = level[cur] + 1;
            }
        }
    }
    return level[destination];
}
void reset() //reset all the arrays and vector for multi test case
{
    for (int i = 0; i < N; i++)
    {
        vis[i] = 0;
        level[i] = 0;
        graph[i].clear();
    }
}
int main()
{
    sieve(); //call the sieve function so that we can use it for all test case hence reduce the time
    int t;
    cin >> t;
    while (t--)
    {
        reset(); // reset all array before using them
        int a, b;
        cin >> a >> b;
        int i;
        vector<int> prime;
        for (i = 1000; i <= N; i++) // inserting all the prime between 1000 to 9999 so that we can make a graph using those
        {
            if (chk[i] == true)
            {
                prime.push_back(i);
            }
        }
        

            int j;
            for (i = 0; i < prime.size(); i++) //generating the graph for all the prime
            {
                for (j = 0; j < prime.size(); j++)
                {
                    if (ok(prime[i], prime[j]) && i != j)
                    {
                        // we will make a edge between any two prime if and only if those prime has one digit difference from each other.
                        graph[prime[i]].push_back(prime[j]);
                    }
                }
            }
            int ans = bfs(a, b);
            cout << ans << '\n';
        
    }
   
}