Longest Common Substring for multiple string

8. Longest Common Substring for multiple string :
                                                Same idea as we used to calculate
   LCS of two string. Just add w(final minimum matching of the state) and
   v(maximum matching of the state of current two string).

Expected Complexity : O(sum(all string)*k), k = total string

const int inf = 1 << 28;
struct state {
    int len;
    int link;
    int w;
    int v;
    int pos;   // to print the string
    map <char, int> next;  // when cause TLE, use array instead of map

    node() {}

    node() {
        len = 0;
        link = -1;
        w = inf;
        v = 0;
        pos = -1;
        next.clear();
    }
} st[mx << 1];

// construct suffix automata
int tNode;
void buildDFA(string s, int len)
{
    init();
    for (int i=0; i<len; i++)
        sz_extend(s[i]);
    tNode = sz;
}

void LCS(string t, int len)
{
    int v = 0;  // initial state
    int l = 0;  // initial match

    for (int i=0; i<len; i++)
    {
        char c = t[i];
        while (v && !st[v].next.count(c))
        {
            v = st[v].link;
            l = st[v].len;
        }
        if (st[v].next.count(c))
        {
            v = st[v].next[c];
            l++;
        }
        st[v].v = max(st[v].v, l);
    }
    for (int i=1; i<tNode; i++)
        st[st[i].link].v = max(st[st[i].link].v, st[i].v);  // copy to root
    for (int i=0; i<tNode; i++)
    {
        st[i].w = min(st[i].w, st[i].v);
        st[i].v = 0;
    }
}

void get_final_len()
{
    int res = -1;
    for (int i=0; i<tNode; i++)
        res = max(res, min(st[i].len, st[i].w);
    cout << "LCS : " << res << endl;
}

Practice Problem :
                 Spoj   : LCS2 - Longest Common Substring II
                 UVA-LA : 3451 - Longest Common Substring