Making Huge Palindromes LightOJ - 1258 (Manacher算法)

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本文介绍了一种算法，用于将任意给定的字符串通过在右侧添加字符的方式转换成尽可能短的回文串。该算法利用了Manacher算法来找出最长的回文子串，并在此基础上构建最小长度的回文串。

Making Huge Palindromes

LightOJ - 1258

A string is said to be a palindrome if it remains same when read backwards. So, 'abba', 'madam' both are palindromes, but 'adam' is not.

Now you are given a non-empty string S, containing only lowercase English letters. The given string may or may not be palindrome. Your task is to make it a palindrome. But you are only allowed to add characters at the right side of the string. And of course you can add any character you want, but the resulting string has to be a palindrome, and the length of the palindrome should be as small as possible.

For example, the string is 'bababa'. You can make many palindromes including

bababababab

babababab

bababab

Since we want a palindrome with minimum length, the solution is 'bababab' cause its length is minimum.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing a string S. You can assume that 1 ≤ length(S) ≤ 10⁶.

Output

For each case, print the case number and the length of the shortest palindrome you can make with S.

Sample Input

bababababa

pqrs

madamimadam

anncbaaababaaa

Sample Output

Case 1: 11

Case 2: 7

Case 3: 11

Case 4: 19

Hint

Dataset is huge, use faster I/O methods.

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 1e6+100;
char s[MAXN];
char News[MAXN<<1];
int len[MAXN<<1];
bool post[MAXN];
int init(int n){
    int i,cnt = 0;
    News[cnt++] = '@';
    News[cnt++] = '#';
    for(i = 0; i < n; i++){
        News[cnt++] = s[i];
        News[cnt++] = '#';
    }
    News[cnt++] = '$';
    News[cnt] = '\0';
    return cnt-1;
}
void Manacher(int n){
    int i,mx = 0,po = 0;
    memset(post,false,sizeof(post));
    for(i = 1; i < n; i++){
        if(mx > i)
            len[i] = min(mx-i,len[2*po-i]);
        else
            len[i] = 1;
        while(News[i-len[i]] == News[i+len[i]])
            len[i]++;
        if(i+len[i] == n)
            post[len[i]-1] = true;
        if(i+len[i] > mx){
            mx = i+len[i];
            po = i;
        }
    }
}
int main(){
    int t,cas = 0;
    scanf("%d",&t);
    while(t--){
        scanf("%s",s);
        int n = strlen(s);
        n = init(n);
        Manacher(n);
        n = strlen(s);
        int i,flag = 0;
        printf("Case %d: ",++cas);
        for(i = n; i >= 2; i--){
            if(post[i]){
                flag = 1;
                printf("%d\n",n+(n-i));
                break;
            }
        }
        if(!flag){
            printf("%d\n",n+n-1);
        }
    }
    return 0;
}