UVa 1401 (Tire tree) Remember the Word

D (i) means the number of decomposition methods of S[i, L-1] from the beginning of the suffix, the string is S[0, L-1]

Then there is d (i) = sum{D (I+len (x)) | Word x is the prefix of S[i, L-1]}

The recursive boundary is D (L) = 1, which represents an empty string.

Constructs a tire tree for the first n words, encounters a word node in the tree to find a suffix that represents the x in a state transition

1#include <cstdio>2#include <cstring>3 4 Const intMaxnode =400000+Ten;5 Const intMAXL = -+Ten;6 Const intSigma_size = -;7 Const intMOD =20071027;8 9 CharS[maxnode];Ten intl; One intD[maxnode]; A CharS1[MAXL]; -  - intsz; the intCh[maxnode][sigma_size]; - intVal[maxnode]; -  -InlineintIdxCharc) {returnC'a'; } +  - voidInit () {sz =1; memset (ch[0],0,sizeof(ch[0])); } +  A voidInsertChar* S,intv) at { -     intn = strlen (s), u =0; -      for(inti =0; I < n; i++) -     { -         intc =idx (s[i]); -         if(!Ch[u][c]) in         { -memset (Ch[sz],0,sizeof(Ch[sz])); toVAL[SZ] =0; +CH[U][C] = sz++; -         } theU =Ch[u][c]; *     } $Val[u] =v;Panax Notoginseng } -  the voidSolve () + { A      for(inti = l1; I >=0; i--) the     { +         intU =0; -          for(intj = i; J < L; J + +) $         { $             intc =idx (s[j]); -             if(!ch[u][c]) Break; -U =Ch[u][c]; the             if(Val[u]) d[i] = (D[i] + d[j +1]) %MOD; -         }Wuyi     } the } -  Wu intN; - intMain () About { $     //freopen ("In.txt", "R", stdin); -  -     intKase =0; -      while(SCANF ("%s", s) = =1) A     { + Init (); theL =strlen (s); -scanf"%d", &n); $          while(n--) {scanf ("%s", S1); Insert (S1,1); }//Val[i] = 1 indicates that this is a word node thememset (D,0,sizeof(d)); theD[L] =1; the solve (); theprintf"Case %d:%d\n", ++kase, d[0] %MOD); -     } in  the     return 0; the}

code June

UVa 1401 (Tire tree) Remember the Word