h3. The first idea is that if you have a string of length h, you can insert it in a trie of height h. We could maintain a list of tries,

h3. trie H[N], where H[i] is a trie with height i so when we insert a string with lenght i, we insert it in H[i].

h3. trie H[N], where H[i] is a trie with height i so when we insert a string with length i, we insert it in H[i].

h3. But what happens when we have, say 2 strings, one of length 3, and one with length 10^20^? We can't declare H[10^20^].

h3. 1) if there is a trie with height h in the binary search tree, we insert the string in that trie

h3. 2) if there isn't a trie with height h in the binary search tree, we create it, we insert the stringh, and then we insert the trie in

h3. 2) if there isn't a trie with height h in the binary search tree, we create it, we insert the string in an empty trie, and then we insert the trie in

h3. the binary search tree.
== code(c) |

#include <cstdio>
#include <string>
#include <set>
#include <vector>
#define maxh 666013
#define N 10024

// Sorting Big Numbers using Trie-Forest based sort indexed in a STL set (Red-Black Tree)

// copyright   idea :          Serban Lupulescu
//             implementation: Mircea Dima

// copyright(c) Serban Lupulescu & Mircea Dima

// Time Complexity: O(n*max_len)
// Faculty of Mathematics and Computer Science, University of Bucharest
// :D

#include <cstdio>
#include <string>
#include <set>
#include <vector>
#define N 10024

using namespace std;

      return n;
}

struct nod { int nr; nod *next[10];};
typedef nod* trie;

{
      for(int i=0; i < n; ++i)
      {

            if(T->next[a[i]] == 0)
            //if I don't have an edge, I creat it

            if(T->next[a[i]] == 0) //if I don't have an edge, I create it

            {
                  T->next[a[i]]=new nod;
                  T->next[a[i]]->nr=0;

            {
                  sol[n]=i;

                  if(T->next[i]->nr)//has a number ended?

                  if(T->next[i]->nr)//has a number ended here?

                  {
                        for(int j=1; j <= T->next[i]->nr; ++j)
                        {