POJ 3261 (suffix array dichotomy) Milk Patterns

This problem and UVA 122,061-like, for at least repeat the longest string of K-times.

First, the length of the longest string, Len, and then segment the height array with Len as the boundary, if there is a section that contains more than k suffixes, it meets the requirements.

1#include <cstdio>2#include <cstring>3#include <algorithm>4 using namespacestd;5 6 Const intMAXN =20000+Ten;7 Const intMAXM =1000000+Ten;8 9 intS[MAXN];Ten intSA[MAXN], HEIGHT[MAXN], RANK[MAXN]; One intT[MAXN], T2[MAXN], C[MAXM]; A intN, K; -  - voidBuild_sa (intm) the { -     intI, *x = t, *y =T2; -      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) A     { at         intp =0; -          for(i = n-k; i < n; i++) y[p++] =i; -          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]; in          for(i = n-1; I >=0; i--) sa[--c[x[y[i] []] =Y[i]; - swap (x, y); top =1; x[sa[0]] =0; +          for(i =1; I < n; i++) -X[sa[i]] = y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+K]? P-1: p++; the         if(P >= N) Break; *m =p; $     }Panax Notoginseng } -  the voidbuild_height () + { A     intI, j, k =0; the      for(i =0; I < n; i++) Rank[sa[i] =i; +      for(i =0; I < n; i++) -     { $         if(k) k--; $j = Sa[rank[i]-1]; -          while(S[i + K] = = S[j + K]) k++; -Height[rank[i]] =K; the     } - }Wuyi  the BOOLOkintlen) - { Wu     intCNT =0; -      for(inti =0; I < n; i++) About     { $         if(i = =0|| Height[i] < len) cnt =0; -         if(++cnt >= k)return true; -     } -     return false; A } +  the intMain () - { $     //freopen ("In.txt", "R", stdin); the  thescanf"%d%d", &n, &k); the      for(inti =0; I < n; i++) the     { -scanf"%d", &s[i]); ins[i]++; the     } theS[n] =0; AboutBuild_sa (1000002); the build_height (); the  the     intL =1, R =N; +      while(L <R) -     { the         intM = (L + R +1) /2;Bayi         if(OK (M)) L =M; the         ElseR = M-1; the     } -  -printf"%d\n", L); the  the     return 0; the}

code June

POJ 3261 (suffix array dichotomy) Milk Patterns