[HDOJ3068]最长回文

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3068

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11870    Accepted Submission(s): 4351

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

 

 

 

Manacher算法的模版题，刚模仿着写了个，趁热用一下

 

 

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>

using namespace std;

const int maxn = 1000010;
int pre[maxn];
char str[maxn];
char tmp[maxn*2];

inline int min(int x, int y) {
    return x < y ? x : y;
}

int init(char *tmp, char *str) {
    int len = strlen(str);
    tmp[0] = '$';
    for(int i = 0; i <= len; i++) {
        tmp[2*i+1] = '#';
        tmp[2*i+2] = str[i];
    }
    tmp[2*len+2] = 0;
    len = 2 * len + 2;
    return len;
}

void manacher(int *pre, char *tmp, int len) {
    int id = 0;
    int mx = 0;
    for(int i = 1; i < len; i++) {
        pre[i] = mx > i ? min(pre[2*id-i], mx-i) : 1;
        while(tmp[i+pre[i]] == tmp[i-pre[i]]) {
            pre[i]++;
        }
        if(pre[i] + i > mx) {
            id = i;
            mx = pre[i] + id;
        }
    }
}

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%s", str)) {
        memset(pre, 0, sizeof(pre));
        memset(tmp, 0, sizeof(tmp));
        int len = init(tmp, str);
        manacher(pre, tmp, len);
        int ans = 1;
        for(int i = 0; i < len; i++) {
            ans = max(ans, pre[i]);
        }
        printf("%d
", ans - 1);
    }
}