hdu 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): 29967    Accepted Submission(s): 10950

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

C/C++:

#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
const int my_max = 3e5;

char my_str[my_max], my_temp[my_max];
int my_ans[my_max], len1, len2;

void manacher()
{
    int id = 0, mx = 0;
    for (int i = 2; i < len2; ++ i)
    {
        if (mx > i) my_ans[i] = min(mx-i, my_ans[id*2 - i]);
        else my_ans[i] = 1;
        while (my_temp[i - my_ans[i]] == my_temp[i + my_ans[i]]) ++ my_ans[i];
        if (my_ans[i] + i > mx)
        {
            mx = my_ans[i] + i;
            id = i;
        }
    }
}

int main()
{
    while (~scanf("%s", my_str))
    {
        len1 = strlen(my_str);
        my_temp[0] = '$';
        my_temp[1] = '#';
        for (int i = 0, j = 2; i < len1; ++ i, j += 2)
        {
            my_temp[j] = my_str[i];
            my_temp[j + 1] = '#';
        }
        len2 = (len1 << 1) + 2;
        my_temp[len2] = '*';

        manacher();
        int temp = 0;
        for (int i = 2; i < len2; ++ i)
            if (temp < my_ans[i])
                temp = my_ans[i];

        printf("%d
", temp - 1);
    }

    return 0;
}