Manacher算法学习资料： http://blog.csdn.net/dyx404514/article/details/42061017

最长回文

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 9282 Accepted Submission(s): 3194

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<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>

using namespace std;

const int N = 110055;

int p[2 * N];//记录回文半径
char str0[N];//原始串
char str[2 * N];//转换后的串

void init()
{
    int i, l;
    str[0] = '@'; str[1] = '#';
    for (i = 0, l = 2; str0[i]; i++, l += 2)
    {
        str[l] = str0[i];
        str[l + 1] = '#';
    }
    str[l] = 0;
}
int solve()
{
    int ans = 0;
    int i, mx, id;
    mx = 0;//mx即为当前计算回文串最右边字符的最大值  
    for (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;//如果i>=mx，要从头开始匹配  
        while (str[i + p[i]] == str[i - p[i]])    
            p[i]++;
        if (i + p[i]>mx)//若新计算的回文串右端点位置大于mx，要更新po和mx的值
        {
            mx = i + p[i];
            id = i;
        }
        ans = max(ans,p[i]);
    }
    return ans - 1;
}
int main()
{
    while (scanf("%s", str0) != -1)
    {
        init();
        printf("%d\n", solve());
    }
    return 0;
}