Spoj Sublex lexicographical Substring Search-suffix array

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Portal II

Main topic

Given a string, ask it multiple times for its $k$ large-nature different substring, outputting it.

Consider suffix trie. Consider each suffix in turn to add the number of essentially different substrings, it is obvious that it is $n-sa[i]-height[i]$.

After finding out the $height$ array, we can find the prefix of the number of substrings of the essence and the two points of each query.

Here you can go directly offline, $O (n + m) $ sweep just fine.

Code

1 /**2 * Spoj3 * Problem#sublex4 * Accepted5 * time:30ms6 * memory:19456k7  */8#include <bits/stdc++.h>9 using namespacestd;TentypedefBOOLBoolean; One  AtypedefclassPair3 { -      Public: -         intx, y, id; the }pair3; -  -typedefclassQuery { -      Public: +         intK, S, T, id; -  +Booleanoperator< (Query b)Const { A             returnK <B.K; at         } - }query; -  - Const intN = 1e5 +5, M =505; -  - intN, M; in CharStr[n], bs[n]; - intRk[n <<1], Sa[n], hei[n], cnt[n]; to Pair3 Ps[n], buf[n]; + Query qs[m]; -  theInlinevoidinit () { *scanf"%s", str +1); $n = strlen (str +1);Panax Notoginsengscanf"%d", &m); -      for(inti =1; I <= m; i++) thescanf"%d", &AMP;QS[I].K), qs[i].id =i; + } A  theInlinevoidRadix_sort (pair3* x, pair3*y) { +     intm = ((n > the) ? (n): ( the)); -memset (CNT,0,sizeof(int) * (M +1)); $      for(inti =1; I <= N; i++) cnt[x[i].y]++; $      for(inti =1; I <= m; i++) Cnt[i] + = cnt[i-1]; -      for(inti =1; I <= N; i++) y[cnt[x[i].y]--] =X[i]; -memset (CNT,0,sizeof(int) * (M +1)); the      for(inti =1; I <= N; i++) cnt[y[i].x]++; -      for(inti =1; I <= m; i++) Cnt[i] + = cnt[i-1];Wuyi      for(inti = n; I i--) x[cnt[y[i].x]--] =Y[i]; the } -  WuInlinevoidBuild_sa () { -      for(inti =1; I <= N; i++) AboutRk[i] =Str[i]; $      for(intK =1, dif =0; K <= N; K <<=1, dif =0) { -          for(inti =1; I <= N; i++) -ps[i].x = Rk[i], ps[i].y = Rk[i + K], ps[i].id =i; - Radix_sort (PS, buf); Ark[ps[1].id] = + +dif; +          for(inti =2; I <= N; i++) theRk[ps[i].id] = (dif + = (ps[i].x! = ps[i-1].x | | Ps[i].y! = ps[i-1].y)); -         if(DIF = =N) $              Break; the     } the      for(inti =1; I <= N; i++) theSa[rk[i]] =i; the } -  inInlinevoidget_height () { the      for(inti =1, J, k =0; I <= N; i++) { the         if(k) k--; About         if(Rk[i] >1) { the              for(j = Sa[rk[i]-1]; i + K <= n && j + k <= n && str[i + K] = = Str[j + K]; k++); theHei[rk[i]] =K; the}Else +hei[1] =0; -     } the }Bayi  theInlinevoidsolve () { the Build_sa (); - get_height (); - //for (int i = 1; I <= n; i++) the //cerr << hei[i] << ""; the //cerr << Endl; theSort (qs +1, QS + M +1); the     Long LongCur =0; -      for(inti =1, NQ =1, Delta; I <= n && nq <= m; i++) { theDelta = n-sa[i] +1-Hei[i]; the         if(cur + Delta <qs[nq].k) theCur + =Delta;94         Else { theQS[NQ].S = Sa[i], qs[nq].t = Sa[i] + hei[i] + (QS[NQ].K-cur); thenq++, i--; the         }98     } About      for(inti =1; I <= m; i++) -          while(Qs[i].id! =i)101 swap (Qs[qs[i].id], qs[i]);102      for(inti =1; I <= m; i++) {103         intLen = qs[i].t-Qs[i].s;104memcpy (BS, str +Qs[i].s, Len); theBs[len] =0;106 puts (BS); 107     }108 }109  the intMain () {111 init (); the solve ();113     return 0; the}

Spoj Sublex lexicographical Substring Search-suffix array