Spoj1811 Longest Common Substring, suffix Automatic Machine

Spoj1811 Longest Common Substring, suffix Automatic Machine
Spoj1811LCS

The longest common substring of two strings.

The procedure is simple. If the match is successful, tl ++ fails to roll back from the parent pointer. tl = t [now]. len.

From this question, we can clearly understand the nature of the fa pointer of the suffix automatic machine:
Point to a status. The acceptance string s [x .. x + I] is an accepted string suffix with the current status s [j-I .. j] matching is the longest one.
Is there a similar nature as KMP?
KMP uses the next array to roll back the data in case of mismatch. the position I to which the data is backed is s [0 .. i] and the current string suffix s [j-I .. j] The longest match.

So.
Suffix automatic machines can be used to obtain the longest matching length between a substring (s [x ..]) and another substring.

KMP can solve the longest matching length between a string (s [0 ..]) and a substring of another string.

#include
 
  #include
  
   #include
   
    #includeusing namespace std;#define Maxn 250100int root,last;//samint tots;struct sam_node{    int fa,son[26];    int len;    void init(int _len){len=_len;fa=-1;memset(son,-1,sizeof(son));}}t[Maxn*2];//length*2void sam_init(){    tots=0;    root=last=0;    t[tots].init(0);}void extend(char ch){    int w=ch-'a';    int p=last;    int np=++tots;t[tots].init(t[p].len+1);    int q,nq;    while(p!=-1&&t[p].son[w]==-1){t[p].son[w]=np;p=t[p].fa;}    if (p==-1) t[np].fa=root;    else{        q=t[p].son[w];        if (t[p].len+1==t[q].len){t[np].fa=q;}        else{            nq=++tots;t[nq].init(0);            t[nq]=t[q];            t[nq].len=t[p].len+1;            t[q].fa=nq;t[np].fa=nq;            while(p!=-1&&t[p].son[w]==q){t[p].son[w]=nq;p=t[p].fa;}        }    }    last=np;}char s[Maxn];char f[Maxn];int work(int l2){    int i,now=root,ind,tl=0;    int ret=0;    for(i=0;i