Ural 1297 suffix array: Find the longest response substring

Train of Thought: This afternoon I started WAF and went through discuss. There is an array in it: abcdefdcba. This output is ABCD, so it is wrong.

Then you will know that the suffix array you have written has a bug with this echo string, and you will not know how to change it.

Then, let's look at the problem solution. rmq is used to pre-process the longest public prefix of any two suffixes first. Because I don't know this, I read it for another afternoon ......

Then, when I understood the combination of rmq and the suffix, I found that rmq is not required here. I just need to handle the above example, haha ...... Wit ......

However, in the papers of the National Training Team, rmq was used as the Final Solution and rmq was used together. It was also used in other topics.

This does not require rmq:

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <map> # include <queue> # include <set> # include <cmath> # include <bitset> # define MEM (, b) memset (a, B, sizeof (A) # define lson I <1, L, mid # define rson I <1 | 1, Mid + 1, R # define llson j <1, L, mid # define rrson j <1 | 1, Mid + 1, R # define INF 0x7fffffff # define maxn 4010 using namespace STD; typedef long ll; typedef unsigned long ull; void Rad IX (int * STR, int * a, int * B, int N, int m) {static int count [maxn]; MEM (count, 0 ); for (INT I = 0; I <n; I ++) ++ count [STR [A [I]; for (INT I = 1; I <= m; I ++) Count [I] + = count [I-1]; for (INT I = n-1; I> = 0; I --) B [-- count [STR [A [I] = A [I];} void suffix (int * STR, int * Sa, int N, int m) {static int rank [maxn], a [maxn], B [maxn]; for (INT I = 0; I <n; I ++) rank [I] = I; radix (STR, rank, SA, n, m); rank [SA [0] = 0; For (INT I = 1; I <n; I ++) Rank [SA [I] = rank [SA [I-1] + (STR [SA [I]! = STR [SA [I-1]); For (INT I = 0; 1 <I <n; I ++) {for (Int J = 0; j <N; j ++) {A [J] = rank [J] + 1; B [J] = J + (1 <I)> = n? 0: rank [J + (1 <I)] + 1; SA [J] = J;} Radix (B, SA, rank, N, N ); radix (A, rank, SA, n, n); rank [SA [0] = 0; For (Int J = 1; j <n; j ++) rank [SA [J] = rank [SA [J-1] + (A [SA [J-1]! = A [SA [J] | B [SA [J-1]! = B [SA [J]) ;}} void calcheight (int * STR, int * Sa, int * H, int N) {static int rank [maxn]; int K = 0; H [0] = 0; For (INT I = 0; I <n; I ++) rank [SA [I] = I; for (INT I = 0; I <n; I ++) {k = 0? 0: K-1; If (rank [I]) while (STR [I + k] = STR [SA [rank [I]-1] + k]) K ++; else K = 0; H [rank [I] = K ;}} int A [maxn], sa [maxn], height [maxn]; int main () {string s; while (CIN> S) {string STR; For (INT I = s. size ()-1; I> = 0; I --) STR + = s [I]; STR = S + "#" + STR; copy (Str. begin (), str. end (), a); int n = Str. size (); suffix (A, SA, N, N + 256); calcheight (A, SA, height, n); int Len = 0, Pos = 1; for (INT I = 1; I <n; I ++) if (SA [I] + SA [I-1] = N-height [I] & ((SA [I] <S. Size ())! = (SA [I-1] <S. size () {If (height [I]> Len) {Len = height [I]; Pos = min (SA [I], sa [I-1]);} else if (height [I] = Len) Pos = min (Pos, min (SA [I], sa [I-1]);} If (LEN> 1) cout <Str. substr (Pos, Len) <Endl; else cout <s [0] <Endl;} return 0 ;} /* abcddcbaefdcbaabcdefdcba // if there is no SA [I] + SA [I-1] = N-height [I] This example will output abcdabcddbcaaaaaaaaa */