Poj 2774 Long Message (suffix array)

Poj 2774 Long Message (suffix array)
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Given two strings A and B, calculate the longest common substring.

Train of Thought: concatenate two strings and separate them with an unused symbol,

Therefore, the answer is to satisfy the height value of the Suffix in different strings with the highest height value.

 

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              # Define eps 1e-6 # define LL long using namespace std; const int maxn = 200000 + 200; // const int INF = 0x3f3f3f; char str1 [100010], str2 [100010]; struct SuffixArray {int s [maxn]; // original character array (the last character must be 0, and the previous character must be not 0) int sa [maxn]; // suffix array. sa [0] Must be n-1, that is, the last character int rank [maxn]; // rank array int height [maxn]; // height array int t [maxn], t2 [maxn], c [maxn]; // auxiliary array int n; // void clear () {n = 0; memset (sa, 0, sizeof (sa);} // m is the maximum character value plus 1 .!!! Set s and n void build_sa (int m) {int I, * x = t, * y = t2; for (I = 0; I <m; I ++) c [I] = 0; for (I = 0; I <n; I ++) c [x [I] = s [I] ++; for (I = 1; I <m; I ++) c [I] + = c [I-1]; for (I = n-1; I> = 0; I --) sa [-- c [x [I] = I; for (int k = 1; k <= n; k <= 1) {int p = 0; for (I = n-k; I <n; I ++) y [p ++] = I; for (I = 0; I <n; I ++) if (sa [I]> = k) y [p ++] = sa [I]-k; for (I = 0; I <m; I ++) c [I] = 0; for (I = 0; I <N; I ++) c [x [y [I] ++; for (I = 0; I <m; I ++) c [I] + = c [I-1]; for (I = n-1; I> = 0; I --) sa [-- c [x [y [I] = y [I]; swap (x, y); p = 1; x [sa [0] = 0; for (I = 1; I <n; I ++) x [sa [I] = y [sa [I-1] = y [sa [I] & y [sa [I-1] + k] = y [sa [I] + k]? P-1: p ++; if (p> = n) break; m = p ;}} void build_height () {int I, j, k = 0; for (I = 0; I <n; I ++) rank [sa [I] = I; for (I = 0; I <n; I ++) {if (k) k --; j = sa [rank [I]-1]; while (s [I + k] = s [j + k]) k ++; height [rank [I] = k ;}} sa; void init () {sa. clear (); int len1 = strlen (str1), len2 = strlen (str2); for (int I = 0; I <len1; I ++) sa. s [I] = str1 [I]-'A' + 1; for (int I = 0; I <len2; I ++) sa. s [len1 + I + 1] = str2 [I]-'A' + 1; sa. s [len1] = 27; sa. s [len1 + len2 + 1] = 0; sa. n = len1 + len2 + 2; sa. build_sa (30); sa. build_height ();} void solve () {int ans = 0; int len1 = strlen (str1); for (int I = 1; I <sa. n; I ++) {if (sa. height [I]> ans) if (sa. sa [I-1]
             
               Len1 | sa. sa [I]
              
                Len1) ans = sa. height [I];} cout <ans <endl;} int main () {// freopen(input.txt, r, stdin); while (scanf (% s, str1, str2) = 2) {init (); solve ();} return 0 ;}