[JSOI2007] Text generator (AC automaton +DP)

Test instructions

Give you a string of N. Ask how many strings of length m make the n strings appear at least once. Output answer membrane 10007 in the sense of the result.

(n<=100, length of each string <=100)

Exercises

Run DP on the AC automaton.

The idea of a tolerant thought, that the number of occurrences at least once was, all possible-the number of times that did not appear at all.

So consider DP, for a string of length I is transferred from I-1, the number of I positions must not be the end of a string in n strings.

So build an AC automaton and run on it. When a son transfers from his father, he must make sure that the point that the son node and the son node reach indirectly through the fail pointer is not the terminating node.

See the code specifically.

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <cmath>5#include <algorithm>6#include <queue>7 using namespacestd;8 Const intMod=10007;9 Const intn=7000;Ten intN,m,tot,ans; One CharS[n]; A intdp[ $][n]; - structac{ -     intnxt[ -],e,fail; the }ac[n]; - voidInsertChars[]) { - //cout<< "Asfsfsdfsdfs" <<endl; -     intlen=strlen (s); +     intnow=0; -      for(intI=0; i<len;i++){ +         if(ac[now].nxt[s[i]-'A']==0) ac[now].nxt[s[i]-'A']=++tot; Anow=ac[now].nxt[s[i]-'A']; at     } -Ac[now].e=1; - } - voidGet_fail () { -queue<int>Q; -      for(intI=0; i<= -; i++){ in         if(ac[0].nxt[i]) { -         //cout<< "SFSD" <<endl; toQ.push (ac[0].nxt[i]); +         } -     } the      while(!Q.empty ()) { *          $         intU=q.front ();//cout<<u<<endl;Panax Notoginseng Q.pop (); -          for(intI=0; i<= -; i++){ the             if(Ac[u].nxt[i]) { +Ac[ac[u].nxt[i]].fail=Ac[ac[u].fail].nxt[i]; Aac[ac[u].nxt[i]].e|=ac[ac[ac[u].fail].nxt[i]].e; the Q.push (Ac[u].nxt[i]); +             } -             Elseac[u].nxt[i]=Ac[ac[u].fail].nxt[i]; $         } $     } - } - intMain () { thescanf"%d%d",&n,&m); -      for(intI=1; i<=n;i++){Wuyiscanf"%s", s); the insert (s); -     } Wu Get_fail (); -dp[0][0]=1; About      for(intI=1; i<=m;i++) $          for(intj=0; j<=tot;j++) -              for(intx=0; x<= -; x + +){ -                 if(ac[ac[j].nxt[x]].e==0) dp[i][ac[j].nxt[x]]= (dp[i][ac[j].nxt[x]]+dp[i-1][J])%MoD; -             } A     intans=1; +      for(intI=1; i<=m;i++){ theans*= -; -ans%=MoD; $     } the      for(intI=0; i<=tot;i++){ theans-=Dp[m][i]; theans= (ans+mod)%MoD; the     } -printf"%d", ans); in     return 0; the}

[JSOI2007] Text generator (AC automaton +DP)