Spoj 220 suffix array: returns the oldest string that appears at least twice and does not overlap.

Idea: N strings are also connected into one, separated by unused characters in the middle, and then the suffix array is obtained.

Because there is no overlap, it is the same as poj 1743, but here there are multiple strings. Each string must determine whether the longest public prefix in it overlaps, so you can save it with an array and then judge it.

# 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 110010 using namespace STD; typedef long ll; typedef unsigned long ull; void R Adix (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) // calculate the suffix array SA {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 (in T 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 * Rank, int N) // obtain the longest public prefix array H {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], rank [maxn]; int V [maxn], Max [15], Min [15], R, N, Len, J; char s [10002]; bool judge (INT m) {MEM (max, 0); MEM (Min, 100); For (INT I = 1; I <Len; I ++) {If (height [I] <m) {MEM (max, 0); MEM (minute, 100); Max [V [SA [I] = sa [I]; min [V [SA [I] = sa [I];} else {max [V [SA [I] = max (MAX [V [SA [I], sa [I]); min [V [SA [I] = min (Min [V [SA [I], sa [I]); max [V [SA [I-1] = max (MAX [V [SA [I-1], sa [I-1]); min [V [SA [I-1] = min (Min [V [SA [I-1], sa [I-1]); For (j = 1; j <= N; j ++) if (MAX [J]-min [J] <m) break; If (j> N) return true ;}} return false ;} int binary () {int ans = 0, mid, L = 0; while (L <= r) {mid = (L + r)> 1; if (Judge (MID) ans = mid, L = Mid + 1; else r = mid-1;} return ans;} int main () {// freopen ("1.txt"," r ", stdin); int t; scanf (" % d ", & T); While (t --) {scanf ("% d", & N); Len = r = J = 0; For (INT I = 1; I <= N; I ++) {scanf ("% s", S); int len1 = strlen (s); If (r <len1> 1) r = len1> 1; // The maximum limit of two points for (int K = 0; k <len1; k ++) A [Len + k] = s [k] + 5, V [Len + k] = I; // the string in which V records a [Len + len1] = ++ J; // The string is separated by characters that have not been used before. V [Len + len1] = 0; Len = Len + len1 + 1;} Len --; suffix (A, SA, Len, 256); calcheight (A, SA, height, rank, Len); printf ("% d \ n", binary ();} return 0 ;}