hdu3068 最长回文

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7867    Accepted Submission(s): 2691

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

 

Recommend

lcy

 

#include<map>
#include<cstdio>
#include <cstring>
#include<string>
using namespace std;
const int maxn = 110010;
int p[maxn << 1],mx,id;
char s[maxn * 2];
char a[maxn * 2];
int longestPalindromeSubstring(char *a,int *p)
{
    memset(p,0,sizeof(p));
    int mx = 0,id = 0,mm = 1;
    int len2 = 1; s[0] = '$';
    for(int i = 0; a[i] != '\0';++i){
             s[len2++] = '#';
             s[len2++] = a[i];
         }
         s[len2++] = '#';
         s[len2++] = '\0';

    for(int i = 1; s[i] != '\0'; ++i){
        p[i] = mx > i ? min(p[2 * id - i],mx - i):1;
        while(s[i + p[i]] == s[i - p[i]]) p[i]++;
        if(i + p[i] > mx){
            mx = i + p[i];
            id = i;
        }
        mm = max(mm,p[i]);
    }
    return mm - 1;
}

int main()
{

	while(~scanf("%s",a))
	{
         int ans = longestPalindromeSubstring(a,p);
         printf("%d\n",ans);
	}
	return 0;
}