hdu 3068 最长回文(manachar模板)

Problem Description

给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度.
回文就是正反读都是一样的字符串,如aba, abba等

 

Input

输入有多组case,不超过120组,每组输入为一行小写英文字符a,b,c...y,z组成的字符串S
两组case之间由空行隔开(该空行不用处理)
字符串长度len <= 110000

 

Output

每一行一个整数x,对应一组case,表示该组case的字符串中所包含的最长回文长度.

 

Sample Input

aaaa

 

abab

 

Sample Output

4

3

 

Source

2009 Multi-University Training Contest 16 - Host by NIT

 

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manachar讲解:   http://blog.csdn.net/bruce_zeng/article/details/8629572

代码:

 #include <vector>
 #include <set>
 #include <algorithm>
 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <cstdlib>
 #include <string>
 #include <cstring>
 #include <queue>
 #include <stack>
 using namespace std;
 #define N 110010*2

 char s[N],str[N];
 int p[N];

 int init()
 {
     int i,j=;
     str[j++]='$';
     for(i=;s[i];i++){
         str[j++]='#';
         str[j++]=s[i];
     }
     str[j++]='#';
     str[j]='\0';
     return j;
 }

 void manachar(int n)
 {
     int mx=,id,i;
     p[]=;
     for(i=;i<n;i++){
         if(mx>i)
             p[i]=min(mx-i,p[*id-i]);
         else
             p[i]=;
         while(str[i-p[i]]==str[i+p[i]])
             p[i]++;
         if(p[i]+i>mx){
             mx=p[i]+i;
             id=i;
         }
     }
 }

 int main()
 {
     int n;
     while(~scanf("%s",s)){
         n=init();
         manachar(n);
         int ans=;
         for(int i=;i<n; i++){
             if(p[i]>ans)
                 ans=p[i];
         }
         printf("%d\n",ans-);
     }
 }