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): 13605Accepted Submission(s): 4947

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
分析：manacher算法的模板题，不多说了。

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
#define MAXN 110010

char s[MAXN],str[MAXN*2];
int p[MAXN*2];
int n;

void init()
{
    int i;
    str[0] = '@'; str[1] = '#';
    for(i=0,n=2; s[i]; i++,n+=2)
    {
        str[n] = s[i];
        str[n+1] = '#';
    }
    str[n] = 0;
}

int solve()
{
    int id;
    int mx=0;
    int ans = 0;
    for(int i=1; str[i]; i++)
    {
        if(mx > i)
            p[i]=p[2*id-i]>(mx-i)?(mx-i):p[2*id-i];
        else
            p[i] = 1;
        while(str[i-p[i]] == str[i+p[i]]) p[i]++;
        if(p[i]+i > mx)
        {
            mx = p[i] + i;
            id = i;
        }
        if(ans < p[i]) ans = p[i];
    }
    return ans-1;
}

int main()
{
    while(scanf("%s",s)!=-1)
    {
        init();
        printf("%d\n",solve());
    }
}
</cstring></cstdio></iostream></span>