Codeforces 427d match & catch (suffix array, string)

Question

 

Reference: http://blog.csdn.net/xiefubao/article/details/24934617

 

Question: give two strings and find the shortest substring. So that the occurrence times of this substring in two strings are equal to 1. The occurrence is defined as overlapping.

 

Solution: Apply the suffix array. Small enumeration length. If a length Len is valid, the SA [I] ranking must exist. Sa [I] And s [I + 1] public prefix length is greater than or equal to Len, and SA [I] and [I-1] public prefix length is less than Len, at the same time, the common prefix lengths of SA [I + 1] and [I + 2] are smaller than Len, and both SA [I] and SA [I + 1] are ensured. The judge function skillfully implements these judgments.

 

 

// If a Len is valid, the SA [I] ranking must exist. // SA [I] And s [I + 1] public prefix length greater than or equal to Len, // and SA [I] and [I-1] public prefix length less than Len, // At the same time, the public prefix length of SA [I + 1] and [I + 2] is smaller than Len, // both SA [I] and SA [I + 1] are supported. # Include <stdio. h> # include <string. h >#include <algorithm> using namespace STD; # define maxn 1000001int wa [maxn], WB [maxn], WV [maxn], WS [maxn]; int CMP (int * r, int A, int B, int L) {return R [a] = R [B] & R [A + L] = R [B + L];} void da (int * r, int * Sa, int N, int m) {int I, j, P, * x = wa, * Y = WB, * t; for (I = 0; I <m; I ++) WS [I] = 0; for (I = 0; I <n; I ++) WS [x [I] = R [I] ++; for (I = 1; I <m; I ++) WS [I] + = ws [I-1]; for (I = n-1; I> = 0; I --) SA [-- ws [x [I] = I; (J = 1, P = 1; P <n; j * = 2, M = P) {for (P = 0, I = N-J; I <N; I ++) y [p ++] = I; for (I = 0; I <n; I ++) if (SA [I]> = J) Y [p ++] = sa [I]-J; for (I = 0; I <n; I ++) WV [I] = x [Y [I]; for (I = 0; I <m; I ++) WS [I] = 0; for (I = 0; I <n; I ++) WS [wv [I] ++; for (I = 1; I <m; I ++) WS [I] + = ws [I-1]; for (I = n-1; I> = 0; I --) sa [-- ws [wv [I] = Y [I]; for (t = x, x = Y, y = T, P = 1, X [SA [0] = 0, I = 1; I <n; I ++) x [SA [I] = CMP (Y, sa [I-1], sa [I], j )? P-1: P ++;} return;} int rankk [maxn], height [maxn]; void calheight (int * r, int * Sa, int N) {int I, j, k = 0; for (I = 1; I <= N; I ++) rankk [SA [I] = I; for (I = 0; I <n; height [rankk [I ++] = k) for (K? K --: 0, j = sa [rankk [I]-1]; R [I + k] = R [J + k]; k ++); return ;} char s [10010]; int len1, len2; int R [10010], sa [10010]; // This is a good method to remove the impact of subscript. Int judge (int n, int K) {int A, B; a = B = 0; For (INT I = 2; I <= N; I ++) {If (height [I] <k) {if (a = 1 & B = 1) return 1; a = 0, B = 0 ;} if (SA [I]> = 0 & SA [I] <len1) A ++; If (SA [I]> len1 & SA [I] <len2) B ++;} return 0;} int main () {scanf ("% s", S); len1 = strlen (s); s [len1] = 1; scanf ("% s", S + len1 + 1); len2 = strlen (s); For (INT I = 0; I <len2; I ++) R [I] = s [I]; R [len2] = 0; DA (R, SA, len2 + 1,128); calheight (R, SA, len2 ); int ans =-1; for (INT I = 1; I <= len1; I ++) {// enumeration length if (Judge (len2 + 1, I )) {ans = I; break;} printf ("% d \ n", ANS); Return 0 ;}

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Codeforces 427d match & catch (suffix array, string)