Find the longest common sub-sequence between two strings: Dynamic Programming Algorithm

//******
// lcs.c
//** - finds the longest common sub-sequence between 2 strings
//** - implements the most famous dynamic programming algorithm
//** - output is a null-terminated string, so must be the input
//** Notes
//** - this package is provided as is with no warranty.
//** - the author is not responsible for any damage caused
//**   either directly or indirectly by using this package.
//** - anybody is free to do whatever he/she wants with this
//**   package as long as this header section is preserved.
//** Created on 2005-01-22 by
//** - Roger Zhang (rogerz@cs.dal.ca)
//** Modifications
//** -
//** Last compiled under Linux with gcc-3
//**************************

#include <stdlib.h>
#include <string.h>

char *lcs(char *s, char *t)
{
   char *result;
   int n = strlen(s), m = strlen(t), i, j, **a;

   if (!n || !m) { /* empty input string */      return "";
   }

   a = (int**)calloc(n + 1, sizeof(int*));
   a[0] = (int*)calloc((n + 1) * (m + 1), sizeof(int));

   for (i = a[0][0] = 0; i <= n; i++) { /* find the length */      if (!i || (a[i] = a[i - 1] + m)) {
         for (j = 0; j <= m; j++) {
            if (!i || !j) { /* initialize the base row/column */               a[i][j] = 0;
            } else if (s[i - 1] == t[j - 1]) { /* diagonal step */               a[i][j] = a[i - 1][j - 1] + 1;
            } else { /* horizontal or vertical step */               a[i][j] = a[i][j - 1] > a[i - 1][j] ? a[i][j - 1] : a[i - 1][j];
            }
         }
      } else {
         abort(); /* memory failure */      }
   }

   if (!(i = a[n][m])) {
      return ""; /* no common sub-sequence */   }

   if (!(result = (char*)malloc(i + 1)) || (result[i] = '\0')) {
      abort(); /* memory allocation failed */   }

   while (n > 0 && m > 0) { /* back track to find the sequence */      if (s[n - 1] == t[m - 1] && m--) {
         result[--i] = s[--n];
      } else {
         a[n][m - 1] >= a[n - 1][m] ? m-- : n--;
      }
   }

   free(a[0]);
   free(a);

   return result;
}