HDU3068(Manacher算法)

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18219    Accepted Submission(s): 6689

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

 

//2016.10.06
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N = 110005;
char str[N], s[N<<1];
int a[N<<1];


//s为待求字符串，len为字符串长度，a[i]为以s[i]为中心的最长回文串的长度的一半。若s为前后添加过特殊字符的字符串，则返回值为原来字符串种最长回文串的长度。
int manacher(char *s, int *a, int len)
{
    a[0] = 0;
    int ans = 0, j;
    for(int i = 0; i < len; )
    {
        while(i-a[i]>0 && s[i+a[i]+1]==s[i-a[i]-1])
              a[i]++;
        if(ans < a[i])ans = a[i];
        j = i+1;
        while(j<=i+a[i] && i-a[i]!=i+i-j-a[i+i-j]){
            a[j] = min(a[i+i-j], i+a[i]-j);
            j++;
        }
        a[j] = max(i+a[i]-j, 0);
        i = j;
    }
    return ans;
}

int main()
{
    int len;
    while(scanf("%s", str)!=EOF)
    {
        len = 2*strlen(str)+1;
        for(int i = 0; str[i] != '\0'; i++)//将字符串每一位中间插入一个特殊字符
        {
            s[i+i] = '\0';
            s[i+i+1] = str[i];
        }
        s[len-1] = '\0';
        printf("%d\n", manacher(s, a, len));
    }

    return 0;
}