String Part 1

What are the string algorithms??? Tire,km,kmp,ac automaton, suffix array, suffix automaton, rk,shift-and/or,manacher .....

™ SO much AH!!!

Also can only slowly learn ...

The next question is the order I do the problem, the difficulty I will ignore (' @@facesymbol@@´)

Bzoj 2434: [NOI2011] Ali's typewriter

The first string is a man of the word!

First build the Actree and Failtree, save the stub point position of each string.

And then from the root of the tree, the roots to the current node of the string is the current typewriter string, while walking off-line query.

For each query (x, y) starting at the end of the s[y] node, the end node of the s[x] with the DFS sequence idea and the tree array, the node of the fail subtree contains several s[y] nodes, which is the answer.

At that time AC automata will not, failtree not, suffix array no, DFS sequence didn't think ... Qaq

1#include <algorithm>2#include <cstdio>3#include <iostream>4#include <cstdlib>5#include <cstring>6#include <vector>7#include <queue>8 9 #defineRep (I, L, R) for (int i = l; I <= R; i++)Ten #defineDown (i, L, R) for (int i = l; I >= R; i--) One #defineMS 123456 A #defineMAX 1037471823 -  - using namespacestd; the  - structnode - {  -     intx, y, N; +     BOOL operator< (ConstNode &k)Const{returnX <k.x;}; -} E[ms];intEC, Fir[ms]; + intC[MS], r[ms][ -], F[ms], H[ms], TC, SS[MS], Ds[ms], Dt[ms], k[ms]; Aqueue<int>p; at intN, m, x, Y, now, NQ; - CharS[ms]; -  - voidDFS (intx) - { -     into = fir[x]; DS[X] = + +Now ; in      while(o) {DFS (E[O].Y); o =E[O].N;} -DT[X] =Now ; to } +  -InlinevoidADD (intXintz) the{into = x; while(o<=tc+1) k[o]+=z, o = o+ (o& (-o)); } *  $InlineintQsum (intx)Panax Notoginseng { -     into = x, ans=0; while(o) ans+=k[o], o = O (o& (-o));  the     returnans; + } A  theInlinevoidQuery (intx) + { -      while(e[nq].x = =x) ${intY=E[NQ].Y; E[nq].y = Qsum (Dt[ss[y])-Qsum (ds[ss[y]]-1); nq++; } $ } -      - intMain () the { -scanf"%s", s);intL=strlen (s); f[0] = -1;WuyiRep (I,0, L-1)  the         if(s[i]!='P'&& s[i]!='B')  -{if(!r[now][s[i]-'a']) tc++, r[now][s[i]-'a']=TC, H[tc]=now, c[tc]=s[i]-'a'; now=r[now][s[i]-'a']; } Wu         Else if(s[i]=='P') Ss[++n] =Now ; -         Elsenow =H[now]; AboutRep (I,0, TC) F[i] =-1; P.push (0); $      while(!p.empty ()) -     { -x =P.front (); P.pop (); -Rep (I,0, -)if(R[x][i]) A         { +             intFO =F[x]; the              while(Fo >=0) -{if(R[fo][i]) Break;ElseFO =F[FO];} $             if(Fo <0) F[r[x][i]] =0;ElseF[r[x][i]] =R[fo][i]; the P.push (R[x][i]); the         } the     } theDown (i, TC,1) E[i].y = i, E[I].N = Fir[f[i]], fir[f[i]] = i; now =0; DFS (0); -Rep (I,1, TC) Fir[i] =0; inscanf"%d", &m); theRep (I,1, M) the{scanf ("%d%d", &x, &y); e[i].x = y, e[i].y = x, E[I].N =i;} AboutSort (e+1, e+1+m); thenow=n=0; nq=1; Rep (I,0, L-1)  the         if(s[i]!='P'&& s[i]!='B') {now=r[now][s[i]-'a']; ADD (Ds[now],1); } the         Else if(s[i]=='P') Query (++n);Else{ADD (Ds[now],-1); now =H[now];} +Rep (I,1, m) e[i].x = E[I].N; Sort (e+1, e+1+M); Rep (I,1, m) printf ("%d\n", e[i].y); -     return 0; the}

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String Part 1