hdu-3068 - longest palindrome (manacher algorithm template)

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 1 /*
 2     Name:hdu-3068-最长回文
 3     Copyright:
 4     Author:
 5     Date: 2018/4/24 16:12:45
 6     Description:
 7     manacher算法模板 
 8 */
 9 #include <iostream>
10 #include <cstdio>
11 #include <string>
12 #include <cstring>
13 #include <math.h>
14 #include <algorithm>
15 using namespace std;
16 const int MAXN = 110005 * 2; // String length*2 
17  
18  void manacher( char str[], int len[], int n){ // Interface 
19      len[ 0 ] = 1 ;
 20      for ( int i = 1 ,j = 0 ; i < (n<< 1 ) - 1 ;++ i){
 21          int p = i >> 1 , q = i - p, r = ((j+ 1 ) >> 1 ) + len[j] - 1 ;
 22          len[i] = r < q? 0:min(r-q+1,len[(j<<1) - i]);
23         while(p > len[i] - 1 && q + len[i] < n && str[p - len[i]] == str[q+len[i]]) ++len[i];
24         if(q + len[i] - 1 > r) j = i;
25     }
26 }
27 struct Solution {
28     string longestPalindrome(string s) {
29         int n = s.size();
30         int len[MAXN];
31         char *str = &s[0];
32         manacher(str,len,n); // Call the interface and get len[] 
33          string tmp = "" ;
 34          int pos = 0 ,max_len = 0 ;
 35          for ( int i = 0 ;i < (n<< 1 ) - 1 ; ++ i){
 36              int tmp_len = (i& 1 )?len[i]<< 1 :(len[i]<< 1 )- 1 ; // Centered on the '#'or character, the string Length is different 
37              if (tmp_len > max_len) pos = i,max_len = tmp_len; //pos records the center point of the target string, max_len represents the string length of the target string (excluding #) 
38              if (i& 1 ) tmp+= " # " ; else tmp+=s[i>> 1 ];     // Make a tmp[0 ..2n-1] string for easy output 
39          }
 40          if (pos& 1 ){    // Find the starting position pos and print length max_len of the string tmp to be printed (for print output) 
41            max_len = (len[pos ] << 2 ) - 1 ;
 42            pos = pos - (len[pos] << 1 ) + 1 ;
 43          }
 44          else {
 45            max_len = (len[pos] << 2 ) - 3 ;
 46             pos = pos - ((len[pos]- 1 )<< 1 );
 47          }
 48          string ans = "" ;
 49          for ( int i = pos, j = 0 ;j < max_len;++ j,++ i){
 50              if (i& 1 ) continue ;
 51              ans+=tmp[i];      // find the character to be printed in tmp, link it up 
52          }
 53          return ans ;
 54      }
 55 };
56 
57 int main()
58 {
59 //    freopen("in.txt", "r", stdin);
60     string str;
61     while (cin >> str) {
62         getchar();
63         Solution tmp;
64         cout<<tmp.longestPalindrome(str).size()<<endl; 
65     }
66     return 0;
67 }