find all distinct palindrome sub-strings

// C++ program to find all distinct palindrome sub-strings
// of a given string
#include <iostream>
#include <map>
using namespace std;

// Function to print all distinct palindrome sub-strings of s
void palindromeSubStrs(string s)
{
    map<string, int> m;
    int n = s.size();

    // table for storing results (2 rows for odd-
    // and even-length palindromes
    int R[2][n+1];

    // Find all sub-string palindromes from the given input
    // string insert 'guards' to iterate easily over s
    s = "@" + s + "#";

    for (int j = 0; j <= 1; j++)
    {
        int rp = 0;   // length of 'palindrome radius'
        R[j][0] = 0;

        int i = 1;
        while (i <= n)
        {
            //  Attempt to expand palindrome centered at i
            while (s[i - rp - 1] == s[i + j + rp])
                rp++;  // Incrementing the length of palindromic
                       // radius as and when we find vaid palindrome

            // Assigning the found palindromic length to odd/even
            // length array
            R[j][i] = rp;
            int k = 1;
            while ((R[j][i - k] != rp - k) && (k < rp))
            {
                R[j][i + k] = min(R[j][i - k],rp - k);
                k++;
            }
            rp = max(rp - k,0);
            i += k;
        }
    }

    // remove 'guards'
    s = s.substr(1, n);

    // Put all obtained palindromes in a hash map to
    // find only distinct palindromess
    m[string(1, s[0])]=1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= 1; j++)
            for (int rp = R[j][i]; rp > 0; rp--)
               m[s.substr(i - rp - 1, 2 * rp + j)]=1;
        m[string(1, s[i])]=1;
    }

    //printing all distinct palindromes from hash map
   cout << "Below are " << m.size()-1
        << " palindrome sub-strings";
   map<string, int>::iterator ii;
   for (ii = m.begin(); ii!=m.end(); ++ii)
      cout << (*ii).first << endl;
}

// Driver program
int main()
{
    palindromeSubStrs("abaaa");
    return 0;
}