Manarcher the longest palindrome substring of a string "record"

Disclaimer: Only the implementation process is written here. Want to learn manacher can see here the implementation of the process, the algorithm involves some of the principles recommended a blog.

It feels very thin.

Introduction: Given a string s, let you find the longest palindrome substring length.

The algorithm roughly implements the process:

one: In order to exclude the effect of the palindrome string length odd or even. A new string of STR is first inserted between every two characters, and the characters that are not present in the original string (# are used here). set P[i] to the maximum radius of the back-text character string centered on the str[i] character. the length of the palindrome string centered on Str[i] in the new string is p[i]-1.

Two: string from front to back p[] array. (Don't ask why before and after)

Three: Enumerate all P[i] values and update the maximum value.

Manacher essence--p[] array. First we have found the previous P[j] value (0<=j<=i-1) when we are seeking p[i]

Preparation of P[i]:

Use MX record max{k+p[K]} (0<=k<=i-1)-the rightmost position in front of which all palindrome strings can be overwritten .

record MX with ID k-- in front of all palindrome strings can be overwritten to the far right when the maximum value is taken The location of the palindrome string centered on the character Str[id] .

(1) According to the previous p[] p[i]

One, mx > i--with Str[id] as the center of the palindrome string to the character str[i] covered.

In this case we can get: p[i]= min (p[2*id-i], mx-i). (why?) Please see my recommended blog)

Two, MX <= i--with Str[id] as the center of the palindrome string at the end of the character str[i] or not overwrite to the character Str[i].

In this case p[i] = 1, because Palindrome has only its own. (why?) Please see my recommended blog)

(2) Of course, the above treatment is not enough, because the processing of the resulting p[i] is not necessarily the optimal palindrome radius we want.

Why Please see my recommended blog)

The follow-up process is well understood.

while (str[i + p[i]] = = str[I-p[i]]) p[i]++;//extends to the left and right until it cannot extend.

(3) Each time the p[i] is found, the ID to be used for the calculation of p[i+1] can be obtained first.

Code implementation:

#include <cstdio> #include <cstring> #include <algorithm> #define MAXN 110100using namespace Std;char s [maxn];//Original string int str[maxn*2];//new string note array size int p[maxn*2];void manacher (char *t) {    int len = strlen (T);    int l = 0;    str[l++] = ' @ ';//Prevent cross-border    str[l++] = ' # ';    for (int i = 0; i < len; i++)    {        str[l++] = t[i];        str[l++] = ' # ';    }    Str[l] = 0;    int mx = 0, id = 0;    int ans = 0;    for (int i = 0; i < l; i++)    {        if (mx > i)//2*id-i is the symmetric point about ID            p[i] = min (p[2*id-i], mx-i);        else            p[i] = 1;        Left and right extension        while (str[i+p[i]] = = Str[i-p[i]]) p[i]++;        if (i + p[i] > mx)//Find calculation p[i+1] Use the ID        {            mx = i + p[i];            id = i;        }        ans = max (p[i]-1, ans);    }    printf ("%d\n", ans);} int main () {while    (scanf ("%s", s)! = EOF)    {        manacher (s);//the longest palindrome substring length of the string s    }    return 0;}

I have the data structure is very slag, the wrong place to welcome correct. (⊙o⊙)

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Manarcher the longest palindrome substring of a string "record"