O (n) palindrome substring (manacher) algorithm

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O (n) palindrome substring (manacher) algorithm

Problem Description:

Enter a string to find the largest palindrome substring in the list. The meaning of a substring is: a string fragment that appears consecutively in the original string. The meaning of a palindrome is: look and look backwards, like Abba and Yyxyy.

Analytical:

The O (n) palindrome string (manacher) algorithm is introduced here

The basic point of the algorithm: first, in a very ingenious way, all possible odd/even length palindrome strings are converted to odd lengths: Insert a special symbol on either side of each character. For example, Abba becomes #a #b#b#a#, ABA becomes #a #b#a#. To further reduce the complexity of the encoding, you can add another special character at the beginning of the string, so that you do not have to deal with cross-border issues specifically, such as \$ #a #b#a#.

The following is an example of string 12212321, which, after the previous step, became s[] = "\$ #1 #2#2#1#2#3#2#1#";

Then use an array of p[i] to record the length (including S[i]) of the longest palindrome string centered on the character S[i], such as the corresponding relationship of S and P:

S # 1 # 2 # 2 # 1 # 2 # 3 # 2 # 1 #
P 1 2 1 2 5 2 1 4 1 2 1 6 1 2 1 2 1
(as you can see, P[i]-1 is exactly the total length of the palindrome string in the original P.S)

The following calculation P[i], the algorithm adds two auxiliary variable ID and MX, where the ID represents the largest palindrome substring center position, MX is id+p[id], that is, the maximum palindrome substring boundary.

The key point of this algorithm is here: if mx > I, then p[i] >= MIN (p[2 * id-i], mx-i).

The specific code is as follows:

`if (mx > i) {      p[i] = (p[2*id-i] < (mx-i) p[2*id-i]: (MX- i));} else1;}      `

When Mx-i > P[j], the palindrome string centered on S[j] is contained in a palindrome string centered on S[id], because I and J are symmetric, the palindrome string centered on s[i] is necessarily contained in a palindrome string centered on S[id], so there must be p[i] = P[j], see.

When P[j] > mx-i, the palindrome string centered on s[j] is not completely contained in a palindrome string centered on S[id], but based on symmetry, it is known that the part surrounded by the two green boxes is the same, that is, the palindrome string centered on the s[i], and its right will expand to the position of MX at least. , which means p[i] >= mx-i. As to whether the post-MX part is symmetrical, it is only one match.

For the case of MX <= i, p[i] can not be made more assumptions, only p[i] = 1, and then to match the

The following gives the original text, further explaining the reason for the linearity of the algorithm

`#include <iostream>#include<cmath>#include<algorithm>#include<string.h>#include<stdio.h>#defineMAXN 110055using namespacestd;Charma[maxn*2];intmp[maxn*2];intl;intManacher (CharS[],intLen) {    intres =0; L=0; Ma[l++] ='\$'; Ma[l++] ='#';  for(intI=0; i<len; i++) {ma[l++] =S[i]; Ma[l++] ='#'; } Ma[l]=0; intMX =0, id =0;  for(intI=0; i<l; i++) {Mp[i]= mx >i min (mp[2*id-i],mx-i):1;  while(Ma[i+mp[i] [= Ma[i-mp[i]]) Mp[i] + +; if(i + mp[i] >mx) {mx= i +Mp[i]; ID=i; }    }}CharS[MAXN];intMain () {//#ifndef Online_judge//freopen ("In.txt", "R", stdin);//#endif//Online_judge     while(~SCANF ("%s", s)) {        intLen =strlen (s);        Manacher (S,len); intAns =0;  for(intI=0; i<l; i++) {ans=Max (ans,mp[i]); } printf ("%d\n", ans-1); }    return 0;}`

O (n) palindrome substring (manacher) algorithm

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