A preliminary study on suffix Tree Group (SA)

There is no any two suffix of LCP these gadgets.

First question: Enter a string s, and the longest string T causes T to appear more than once in S. The length of the output T.

1#include <algorithm>2#include <stdio.h>3#include <string.h>4 #defineN 1005 Chars[n+1];6 intN, Sa[n], rank[n], height[n];7 namespaceSabuilder {8     intCnt[n], t1[n<<1], t2[n<<1];9     voidWork () {Ten         intI, k, m = -, p, *x = t1, *y =T2; Onememset (CNT,0, M *sizeof(int)); A          for(i =0; I < n; ++i) ++cnt[x[i] = s[i]-'a']; -          for(i =1; I < m; ++i) Cnt[i] + = cnt[i-1]; -          for(i = n-1; I >=0; -i) sa[--cnt[x[i]] =i; the          for(k =1; K <= N; K <<=1) { -              for(i = n-k, p =0; I < n; ++i) y[p++] =i; -              for(i =0; I < n; ++i)if(Sa[i] >= k) y[p++] = Sa[i]-K; -memset (CNT,0, M *sizeof(int)); +              for(i =0; I < n; ++i) + +Cnt[x[y[i]]; -              for(i =1; I < m; ++i) Cnt[i] + = cnt[i-1]; +              for(i = n-1; I >=0; -i) sa[--cnt[x[y[i] []] =Y[i]; A Std::swap (x, y); atx[sa[0]] =0; -              for(i = m =1; I < n; ++i) { -X[sa[i]] = y[sa[i]]==y[sa[i-1]]&&sa[i]+k<n&&sa[i-1]+k<n&&y[sa[i]+k]==y[sa[i-1]+K]? M-1: m++; -             } -             if(M > N) Break; -         } inmemcpy (rank, x, n *sizeof(int)); -          for(i =0, p =0; I < n; ++i)if(Rank[i]) { to             if(p)--p; +K = sa[rank[i]-1]; -              while(S[i+p] = = S[k+p]) + +p; theHeight[rank[i]] =p; *         } $     }Panax Notoginseng } - intMain () { theRegisterintI, ans =0; + gets (s); An =strlen (s); the sabuilder::work (); +printf"%d", ans); -     return 0; $}

A preliminary study on suffix Tree Group (SA)