KMP basic questions. I wrote it for a long time because I didn't have a clear idea at the beginning... Sad reminder.
To clarify the meaning of the question, the question is centered on KMP.AlgorithmThe step for matching your own strings
Suppose there are two strings "S" and "T". We need to determine whether "S" exists in "T.
The Save [I] I used indicates that the save [J]1 after the I position is equal to the Save [I]1 before the S string, then, based on this nature, we can solve the problem.
Stringmatching Automata
Time limit:1000 ms 

Memory limit:65536 K 
Total submissions:113 

Accepted:70 
Description
The finite state automaton (FSA) is an important model of behavioral investors in computer science, linguistics and other areas. a fsa can be typically modeled as a string pattern recognizer described by a quintuple <Σ, S, S0, Delta, F>, where:
Σ is the input alphabet (a finite nonempty set of symbols ).
S is a finite nonempty set of States.
S0 is an element in S designated as the initial state.
Delta is a function delta: S x Σ → S known as the transition function.
F is a (possibly empty) subset of s whose elements are designated as the final states.
An FSA with the above description operates as follows:
At the beginning, the automaton starts in the initial state s0.
The automaton continuously reads symbols from its input, one symbol at a time, and transits between States according to the transition function delta. to be specific, let s be the current state and W the symbol just read, the automation on moves to the state given by Delta (S, W ).
When the automaton reaches the end of the input, if the current state belongs to F, the string consisting sequentially of the symbols read by the automaton is declared accepted, otherwise it is declared rejected.
Just as the name implies, a stringmatching automaton is a FSA that is used for string matching and is very efficient: they examine each character exactly once, taking constant time per text character. the matching time used (after the automation is built) is therefore seconds (n ). however, the time to build the automaton can be large.
Precisely, there is a stringmatching Automaton for every pattern P That you search for in a given text string T. for a given pattern of length m, the corresponding automaton has (m + 1) States {q0, Q1 ,..., QM}: q0 is the start state, Qm is the only final state, and for each I in {0, 1 ,..., M}, if the automaton reaches State Qi, it means the length of the longest prefix of P that is also a suffix of the input string is I. when we reaches state Qm, it means P is a suffix of the currently input string, which suggest we find an occurrence of P.
The following graph shows a stringmatching Automaton for the pattern "ababaca", and how the automaton works given an input string "ababacaba ".
Apparently, the matching process using stringmatching automata is quite simple (also efficient). However, building the automation on efficiently seems to be tough, and that's your task in this problem.
Input
Several lines, each line has one pattern consist of only lowercase alphabetic characters. The length of the longest pattern is 10000. The input ends with a separate line of '0 '.
Output
For each pattern, output shoshould contain (m + 1) lines (M is the length of the pattern ). the Nth line describes how the automaton changes its state from the staten1 after reading a character. it starts with the State number (n1), and then 26 State numbers follow. the 1st State number P1 indicates that when the automaton is in staten1, it will transit to state P1 after reading a character 'a '. the 2N D State number P2 indicates that when the automaton is in staten1, it will transit to state P2 after reading a character 'B '... And so on.
Sample Input
Ababaca0
Sample output
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 02 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 03 1 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 04 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 07 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Source
2010 national programming Invitational contest host by zstu
# Include <stdio. h> # Include < String . H> # Include <Iostream> Using Namespace STD; Char G [ 10010 ]; Int Ans [ 10010 ] [ 30 ]; Int Save [ 10010 ]; Int Main (){ While (Scanf ( " % S " , G) & G [ 0 ]! = ' 0 ' ){ Int Len = Strlen (g); memset (ANS, 0 , Sizeof (ANS); memset (save, 0 , Sizeof (SAVE); ans [ 0 ] [G [0 ] ' A ' ] = 1 ; Int J = 0 ; Int CNT = 1 ; For ( Int I = 1 ; I <= Len; I ++, CNT ++){ If (I! = Len) ans [CNT] [G [I]  ' A ' ] = (I + 1 ); Ans [CNT] [G [ 0 ] ' A ' ] = Max ( 1 , ANS [CNT] [G [ 0 ]' A ' ]); Int TJ = I 1 ; While ( 1 ){ If (TJ <= 0 ) Break ; Ans [CNT] [G [save [TJ]  ' A ' ] = Max (ANS [CNT] [G [save [TJ] ' A ' ], Save [TJ] + 1 ); TJ = Save [TJ] 1 ;} If (I = Len) Break ; /////////////////////////// // While (G [I]! = G [J] & J! = 0 ) {J = Save [J 1 ];} If (G [I] = G [J]) {J ++ ;} Save [I] = J; //////////// // } For (Int I = 0 ; I <= Len; I ++ ) {Printf ( " % D " , I ); For ( Int J = 0 ; J < 26 ; J ++ ) Printf ( " % D " , ANS [I] [J]); printf ( " \ N " );}} Return 0 ;}