"BZOJ1212" L language (AC automata)

"BZOJ1212" L language (AC automaton) surface

Bzoj

Exercises

Naturally, since you want to match the word, it's all thrown into the ac\ automaton.

Now think about how to match.
First \ (ac\) the normal match of the automaton
If this position is able to match the previous string
We need to figure out how this string is covered in this text string.
Whether its previous bit is exactly covered
If there is.
Nor can we directly calculate
Because it could be a different kind of string.
So, just open an array differential
Indicates that the current position can match

Finally, for each text string
Sweep past the difference score Group
Until there is \ (0\) the direct output is OK

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <set>#include <map>#include <vector>#include <queue>using namespaceStdinline intRead () {intx=0, t=1;CharCh=getchar (); while((ch<' 0 '|| Ch>' 9 ') &&ch!='-') Ch=getchar ();if(ch=='-') t=-1, Ch=getchar (); while(ch<=' 9 '&&ch>=' 0 ') x=x*Ten+ch-48, Ch=getchar ();returnX*t;}structnode{intvis[ -];intDep,lst,fail;} t[ the];intN,m,tot;Chars[2000000];intcc[2000000];BOOLvis[2000000];voidInsertChar*s) {intL=strlen (S+1), now=0; for(intI=1; i<=l;++i) {if(!t[now].vis[s[i]-97]) T[now].vis[s[i]-97]=++tot; Now=t[now].vis[s[i]-97];    T[now].dep=i; } t[now].lst=1;}voidGetfail () {queue<int> Q; for(intI=0;i< -; ++i)if(t[0].vis[i]) Q.push (t[0].vis[i]); while(! Q.empty ()) {intU=q.front (); Q.pop (); for(intI=0;i< -; ++i)if(T[u].vis[i]) T[t[u].vis[i]].fail=t[t[u].fail].vis[i],q.push (t[u].vis[i]);ElseT[u].vis[i]=t[t[u].fail].vis[i]; }}intMain () {n=read (); M=read (); for(intI=1; i<=n;++i) {scanf ("%s", S+1);    Insert (s); } getfail (); while(m--) {scanf ("%s", S+1);intL=strlen (S+1);intlst=0, now=0; Memset (CC,0,sizeof(cc)); memset (Vis,0,sizeof(VIS)); vis[0]=true; for(intI=1; i<=l;++i) {now=t[now].vis[s[i]-97]; for(intJ=now;j;j=t[j].fail)if(T[J].LST&AMP;&AMP;VIS[I-T[J].DEP]) {vis[i]=true; Cc[i-t[j].dep+1]++; Cc[i+1]--; }        } for(intI=1, tot=0; i<=l;++i) {tot+=cc[i];if(tot) ++lst;Else  Break; } printf ("%d\n", LST); }return 0;}

"BZOJ1212" L language (AC automata)