hdu3068（最长回文manacher）

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7247    Accepted Submission(s): 2479

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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#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define MIN(a,b) a>b?b:a
using namespace std;
const int MAXN=111000;
char a[MAXN],b[MAXN*2+10];
int len[MAXN*2+10];

void solve(int n)
{
	int i;
	memset(b,'#',sizeof(b));
	b[0]='@',b[2*n+2]='$';
	for(i=0;i<n;i++)
		b[i*2+2]=a[i];
}
int manacher(int n)
{
	int i,id,mx=0;
	for(i=1;i<n-1;i++)
	{
		if(mx>i) len[i]=MIN(len[2*id-i],mx-i);
		else len[i]=1;
		for(;b[i+len[i]]==b[i-len[i]];len[i]++)
			;
		if(i+len[i]>mx)
		{
			mx=i+len[i];
			id=i;
		}
	}
	int max;
	for(max=i=1;i<n-1;i++)
		if(len[i]>max)max=len[i];
	return max-1;	
}
int main()
{
	while(~scanf("%s",&a))
	{
		int n=strlen(a);
		solve(n);
		cout<<manacher(2*n+3)<<endl;
	}
	return 0;
}