Manacher

[hdu 3068|http://acm.hdu.edu.cn/showproblem.php?pid=3068]

• Algorithm:
  https://segmentfault.com/a/1190000003914228

const int MaxN = 1000000 + 7;
char ms[MaxN * 2];
int mr[MaxN * 2];
int manacher( char *s ) {
  int len = strlen( s ), l = 0;
  ms[l++] = '$';
  for( int i = 0; i < len; ++i, l += 2 ) {
    ms[l] = '#', ms[l + 1] = s[i];
  }
  ms[l++] = '#';

  int mx = 0, mp = 0, ans = 0;
  for( int i = 0; i < l; ++i ) {
    mr[i] = mx > i ? min( mr[mp * 2 - i], mx - i ) : 1;
    while( ms[i + mr[i]] == ms[i - mr[i]] )  mr[i]++;
    if( i + mr[i] > mx ) {
      mx = i + mr[i], mp = i;
    }
    ans = max( ans, mr[i] - 1 );
  }

  return ans;
}

• Note

  1. 填充首末间隙，并且首部再加一个（使整个串S是回文串时失配）\

  mr[i] = mx > i ? min( mr[mp * 2 - i], mx - i ) : 1;
  while( ms[i + mr[i]] == ms[i - mr[i]] )  mr[i]++;
  if( i + mr[i] > mx ) mx = i + mr[i], mp = i;

3. 类似尺取，S_i 只进一次 => O(n)