"SAM" codevs3160-the longest common child string

"The main topic"

The longest common substring of a two string.

Ideas

Establish a suffix automaton on the first string, and a second string to match. CNT records the length of the current longest common substring, while RET records the answer.

P represents the position pointer, initially in the RT position.

For one s[i of the second string], if there is currently s[i] child, then cnt+1, continue to move backwards, otherwise along the pre pointer. If the pre pointer returns to 0, then p back to rt,cnt is emptied to 0, otherwise if the middle has a little s[i] child, cnt=step[]+1.

Why cnt=step[]+1? Do not forget that the essence of the suffix automaton is to maintain the suffix, running along the pre pointer to a smaller length of the suffix to move, a position represented by the suffix of the longest length of step[], plus s[i], that is step[]+1.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5#include <cmath>6 using namespacestd;7 Const intmaxn=100000+5;8 intN;9 Charstr[2][MAXN];Ten structSAM One { A     intstep[maxn*2],pre[maxn*2],next[maxn*2][ -]; -     intTot,last; -InlineintNewNode (intCNT) the     { -step[++tot]=CNT; -pre[tot]=0; -          for(intI=0;i< -; i++) next[tot][i]=0; +         returntot; -     } +      AInlinevoidExtendintx) at     { -         intp=Last ; -         intNp=newnode (step[p]+1); -          while(P &&!next[p][x]) next[p][x]=np,p=Pre[p]; -         if(!p) pre[np]=1; -         Else in         { -             intq=Next[p][x]; to             if(step[q]==step[p]+1) pre[np]=Q; +             Else  -             { the                 intNq=newnode (step[p]+1); *                  for(intI=0;i< -; i++) next[nq][i]=Next[q][i]; $pre[nq]=Pre[q];Panax Notoginsengpre[q]=pre[np]=NQ; -                  while(p&&next[p][x]==q) next[p][x]=nq,p=Pre[p]; the             } +      A         } thelast=NP; +     } -      $InlinevoidClear () $     { -         inttot=0; -last=NewNode (tot); the     } -     WuyiInlineintQuery () the     {   -         intret=0, cnt=0;  Wu         intp=1;  -          for(intI=0; str[1][i];i++) About         {   $             intindex=str[1][i]-'a'; -             if(Next[p][index]) p=next[p][index],cnt++; -             Else -             {     A                  while(P &&!next[p][index]) p=Pre[p]; +                 if(!p) p=1, cnt=0;  the                     Elsecnt=step[p]+1, p=Next[p][index]; -                     /*since the string that is returned along the pre is the suffix of the current string, the first child with index is the longest-satisfying suffix, which is the length of step+1*/   $             } theret=Max (ret,cnt);  the         }   the         returnret;  the     }   - }suf; in  the voidInit () the { Aboutscanf"%d",&n); thescanf"%s", str[0]); the     intLen=strlen (str[0]); the suf.clear (); +      for(intI=0; i<len;i++) Suf.extend (str[0][i]-'a'); -scanf"%s", str[1]); the }Bayi  the intMain () the { - init (); -printf"%d", Suf. Query ()); the     return 0; the}

"SAM" codevs3160-the longest common child string