杭电3068 最长回文 最长回文的manacher算法

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12275    Accepted Submission(s): 4514

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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我就不详解了看这里 http://blog.youkuaiyun.com/xingyeyongheng/article/details/9310555

附ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s[330000];
int i,j,k,l,p[330000],m,n,id,Maxlen;
int main()
{
	while(scanf("%s",s)!=EOF)
	{
		l=strlen(s);
		for(i=l;i>=0;i--)
		{
			s[2*i+1]='#';
			s[2*i+2]=s[i];
		}
		id=0;
		Maxlen=0;
		s[0]='*';
		for(i=2;i<2*l+1;i++)
		{
			if(p[id]+id>i)
			p[i]=min(p[2*id-i],p[id]+id-i);
			else
			p[i]=1;
			while(s[i-p[i]]==s[i+p[i]])
			p[i]++;
			if(id+p[id]<i+p[i])
			id=i;
			if(Maxlen<p[i])
			Maxlen=p[i];
		}
		printf("%d\n",Maxlen-1);
	}
}