Spoj694 -- suffix Array

Address: http://www.spoj.com/problems/DISUBSTR/

Question meaning:

Here is a string to remove the number of different substrings.

Solution:

Each substring must be the prefix of a suffix, so the original problem is equivalent to finding out the inconsistency between all suffixes.
The number of prefixes. If all suffixes follow suffix (SA [1]), suffix (SA [2]),
Suffix (SA [3]),... , Suffix (SA [N]), it is not difficult to find that
Suffix (SA [k]), which generates N-sa [k] + 1 new prefix. However
Height [k] is the same as the prefix of the preceding string. So suffix (SA [k]) will "Contribute"
Returns N-sa [k] + 1-height [k] Different substrings. The accumulated answer is the answer to the original question. This practice
The time complexity is O (n ).

Code:

# Include <cstdio> # include <cstring> # include <algorithm> using namespace STD; const int maxn = 1000 + 10; char STR [maxn]; int wa [maxn], WB [maxn], WV [maxn]; int C [maxn]; int CMP (int * r, int A, int B, int L) {return R [a] = R [B] & R [A + L] = R [B + L];} void cal_the_houzhui (char * r, int * Sa, int N, int m) {int I, j, P; int * x = wa, * Y = WB; // sort the first character of each Suffix in the base order for (I = 0; I <m; I ++) C [I] = 0; for (I = 0; I <n; I ++) C [x [I] = R [I] ++; for (I = 1; I <m; I + +) C [I] + = C [I-1]; for (I = n-1; I> = 0; I --) sa [-- C [x [I] = I; // then sort the log-level base to make the SA array stable p = 1; for (j = 1; P <n; j * = 2, M = P) {// base sorting for the second dimension // you can use the SA array directly to sort the result. // y [p] indicates that the result is in the two-dimensional sorting, where is the first position of the smallest p? For (P = 0, I = N-J; I <n; I ++) y [p ++] = I; // from N-J ~ N-1 is impossible to have the second-dimensional, because the second-dimensional has exceeded n-1 //, so they are the smallest in the second-dimensional sorting, of course, it is the first for (I = 0; I <n; I ++) if (SA [I]> = J) Y [p ++] = sa [I]-J; for (I = 0; I <n; I ++) // re-define WV [I] = x [Y [I] According to the Y array; for (I = 0; I <m; I ++) c [I] = 0; for (I = 0; I <n; I ++) C [wv [I] ++; for (I = 1; I <m; I ++) C [I] + = C [I-1]; for (I = n-1; I> = 0; I --) sa [-- C [wv [I] = Y [I]; swap (x, y); P = 1; X [SA [0] = 0; for (I = 1; I <n; I ++) x [SA [I] = CMP (Y, sa [I-1], sa [I], j )? P-1: P ++ ;}} int rank [maxn], height [maxn]; void cal_the_height (char * r, int * Sa, int N) {int I, j, k = 0; // obtain rank for (I = 0; I <n; I ++) rank [SA [I] = I; for (I = 0; I <n; height [rank [I ++] = k) {for (K? K --: 0, j = sa [rank [I]-1]; R [I + k] = R [J + k];) K ++ ;}} int N; int _ SA [maxn]; void solve () {cal_the_houzhui (STR, _ SA, N + 1,256); cal_the_height (STR, _ SA, n + 1 ); long long ans = 0; For (INT I = 1; I <= N; I ++) {ans + = N-_ SA [I]-height [I];} printf ("% LLD \ n", ANS) ;}int main () {int t; scanf ("% d", & T); While (t --) {scanf ("% s", STR); n = strlen (STR); solve ();} return 0 ;}