POJ 3693 (suffix array) Maximum repetition substring

Find the most repeating string, and if there are multiple solutions, ask for the smallest dictionary order.

I also followed the paper Ro chicory to brush this problem.

The first is to enumerate the length of a circular section L, if it appears two times, will certainly contain s[0], s[l], s[2l] these adjacent two.

Then enumerate the adjacent two, extending forward and backward as far as possible, assuming that the extension length is K, the number of repetitions is K/L + 1

Backward extension is very natural is to seek the LCP, this with RMQ pretreatment can be O (1) query.

The forward extension is to consider that the s[i*l] and s[(i+1) *l] that we enumerate are not necessarily the beginning of the string, so I-(k-i) positions are moved forward.

Why move forward so much, because of this, the length of the LCP is exactly the number of I multiples.

So move forward, then ask for LCP again, see if I-(k% i) so much, enough words, the number of repetitions of this string plus 1.

As for the smallest dictionary order, you have to use the height array to naturally bring the dictionary order. Record the maximum length of all repetitions and then enumerate by dictionary order, and output as soon as one group is found.

1#include <cstdio>2#include <cstring>3#include <algorithm>4 using namespacestd;5 6 Const intMAXN =100000+Ten;7 CharS[MAXN];8 intN;9 intSA[MAXN], RANK[MAXN], HEIGHT[MAXN];Ten intT[MAXN], T2[MAXN], c[ the]; One  A voidBuild_sa (intNintm) - { -     intI, *x = t, *y =T2; the      for(i =0; I < m; i++) C[i] =0; -      for(i =0; I < n; i++) C[x[i] = s[i]]++; -      for(i =1; I < m; i++) C[i] + = c[i-1]; -      for(i = n-1; I >=0; i--) Sa[--c[x[i]] =i; +      for(intK =1; K <= N; K <<=1) -     { +         intp =0; A          for(i = n-k; i < n; i++) y[p++] =i; at          for(i =0; I < n; i++)if(Sa[i] >= k) y[p++] = Sa[i]-K; -          for(i =0; I < m; i++) C[i] =0; -          for(i =0; I < n; i++) c[x[y[i]]]++; -          for(i =1; I < m; i++) C[i] + = c[i-1]; -          for(i = n-1; I >=0; i--) sa[--c[x[y[i] []] =Y[i]; - swap (x, y); inp =1; x[sa[0]] =0; -          for(i =1; I < n; i++) toX[sa[i]] = y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+K]? P-1: p++; +         if(P >= N) Break; -m =p; the     } * } $ Panax Notoginseng voidbuild_height () - { the     intK =0; +      for(inti =1; I <= N; i++) Rank[sa[i] =i; A      for(inti =0; I < n; i++) the     { +         if(k) k--; -         intj = Sa[rank[i]-1]; $          while(S[i + K] = = S[j + K]) k++; $Height[rank[i]] =K; -     } - } the  - intd[maxn][ -];Wuyi  the voidINIT_RMQ () - { Wu      for(inti =0; I < n; i++) d[i][0] = Height[i +1]; -      for(intj =1; (1<< j) <= N; J + +) About          for(inti =0; i + (1<< j)-1< n; i++) $D[i][j] = min (d[i][j-1], D[i + (1<< (J-1))][j-1]); - } -  - intRMQ (intLintR) A { +     intK =0; the      while( (1<< (k +1)) <= (R-l +1)) k++; -     returnMin (D[l][k], d[r-(1&LT;&LT;K) +1][k]); $ } the  the intLCP (intIintj) the { thei = Rank[i]-1; j = Rank[j]-1; -     if(I >j) Swap (I, j); in     returnRMQ (i +1, j); the } the  About intA[MAXN], CNT, MAXL; the  the intMain () the { +     //freopen ("In.txt", "R", stdin); -  the     intKase =0;Bayi      while(SCANF ("%s", s) = =1&& s[0] !='#') the     { then =strlen (s); -BUILD_SA (n +1, the); - build_height (); the init_rmq (); the  theCNT = MAXL =0; the          for(inti =1; I < n; i++) -         { the              for(intj =0; j + i < n; J + =i) the             { the                 intK = LCP (J, J +i);94                 intt = k/i +1; the                 intleft = i-(k%i); the                 intHead = J-Left ; the                 if(Head >=0&& LCP (head, head + i) >= left) t++;98                 if(T >Maxl) About                 { -CNT =0;101a[cnt++] =i;102MAXL =T;103                 }104                 if(t = = Maxl) a[cnt++] =i; the             }106         }107 108         intLen =-1, St;109          for(inti =1; I <= n && len = =-1; i++) the         {111              for(intj =0; J < CNT; J + +) the             {113                 intL =A[j]; the                 if(LCP (Sa[i], Sa[i] + L) >= (MAXL-1) *l) the                 { theLen =l;117St =Sa[i];118                      Break;119                 } -             }121         }122 123printf"Case %d:", ++Kase);124          for(inti =0; I < Len * MAXL; i++) printf ("%c", S[st +i]); thePuts"");126     }127  -     return 0;129}

code June

POJ 3693 (suffix array) Maximum repetition substring