Spoj 8222 substrings (SAM)

indicate the source for reprint, thank you http://blog.csdn.net/ACM_cxlove? Viewmode = contents
by --- cxlove

Give a stringS, OrderF (x)IndicatesSIsXThe maximum number of occurrences of the substring. PleaseF (1) .. f (length (s )) (Thanks for CLJ's translation >_<)

Http://www.spoj.pl/problems/NSUBSTR/

When a Sam is created, the Len value of the node indicates the suffix length at that time. Although it is not necessarily a suffix, it can represent the length of a substring.

Starting from the current status, the number of paths that can reach the final state is the number of occurrences of the current length substring, and the maximum value is enough.

The method is to traverse the node topology on the Sam from the back to update the parent node from the child node (the pre on the Sam is not actually the parent node ).

The last case is that the sub-string is completely contained, that is, the long sub-string contains various short sub-strings.

The step of topology is to look at others, and the spoj time limit is too tight. Use length for a ing.

I still don't know much about Sam, but I can accept nodes with suffixes.

The simplest way is to traverse all the suffixes of a string.

# Include <iostream> # include <cstdio> # include <map> # include <cstring> # include <cmath> # include <vector> # include <algorithm> # include <set> # include <string> # include <queue> # define INF 100000005 # define M 40 # define n 510005 # define maxn 300005 # define EPS 1e-10 # define zero () FABS (a) <EPS # define min (A, B) (a) <(B )? (A) :( B) # define max (A, B) (a)> (B )? (A) :( B) # define Pb (a) push_back (a) # define MP (a, B) make_pair (a, B) # define MEM (A, B) memset (a, B, sizeof ()) # define ll unsigned long # define mod 1000000007 # define lson step <1 # define rson step <1 | 1 # define sqr (A) () * (a) # define key_value ch [CH [root] [1] [0] # define test puts ("OK"); # pragma comment (linker, "/Stack: 10241000000,1024000000") using namespace STD; struct Sam {Sam * Pre, * son [26]; int Len, G;} que [n ], * Root, * tail, * B [N]; int tot; void add (INT C, int L) {Sam * P = tail, * NP = & que [tot ++]; NP-> Len = L; tail = NP; while (P & P-> son [c] = NULL) p-> son [c] = NP, P = p-> pre; If (P = NULL) NP-> pre = root; else {Sam * q = p-> son [c]; If (p-> Len + 1 = Q-> Len) NP-> pre = Q; else {Sam * NQ = & que [tot ++]; * NQ = * q; NQ-> Len = p-> Len + 1; NP-> pre = Q-> pre = NQ; while (P & P-> son [c] = q) P-> son [c] = NQ, P = p-> pre; }}} char STR [n/2]; int DP [n/2]; int main () {While (SC ANF ("% s", STR )! = EOF) {int n = strlen (STR); Tot = 0; root = tail = & que [tot ++]; for (INT I = 0; I <N; I ++) add (STR [I]-'A', I + 1); int CNT [n/2]; MEM (CNT, 0 ); for (INT I = 0; I <tot; I ++) CNT [que [I]. len] ++; For (INT I = 1; I <= N; I ++) CNT [I] + = CNT [I-1]; for (INT I = 0; I <tot; I ++) B [-- CNT [que [I]. len] = & que [I]; for (INT I = 0; I <n; I ++) (Root = root-> son [STR [I]-'a'])-> G ++; MEM (DP, 0); For (INT I = tot-1; i> 0; I --) {DP [B [I]-> Len] = max (DP [B [I]-> Len], B [I]-> G); If (B [I]-> pre) B [I]-> pre-> G + = B [I]-> G ;} for (INT I = n-1; I> 0; I --) DP [I] = max (DP [I], DP [I + 1]); for (INT I = 1; I <= N; I ++) printf ("% d \ n", DP [I]);} return 0 ;}