hdoj 3068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): 12270    Accepted Submission(s): 4511

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 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N = 110005;
char s[N],a[N*2];
int p[N*2];
int len;
void manacher(char *s)
{
	int l = 0;
	a[l++] = '$';
	a[l++] = '#';
	for(int i = 0; i < len; i++)
	{
		a[l++] = s[i];
		a[l++] = '#';
	}
	a[l] = 0;
	int id = 0, ml = 0;
	for(int i = 0; i < 2*len+2; i++)
	{
		if(p[id]+id > i)
			p[i] = min(p[2*id-i], p[id]+id-i);
		else
			p[i] = 1;
		while(a[i+p[i]] == a[i-p[i]])
			p[i]++;
		if(id+p[id] < i+p[i])	
			id = i;
		if(ml < p[i])	
			ml = p[i];
	}
	printf("%d\n", ml-1);
} 

int main()
{
	while(scanf("%s",s)!=EOF)
	{
		len = strlen(s);
		manacher(s);
	}
	return 0;
 }