LightOJ 1258 Making Huge Palindromes 【Manacher算法】

题目链接：传送

Making Huge Palindromes

Time Limit: 1000MSMemory Limit: 32768KB64bit IO Format: %lld & %llu

Submit Status uDebug

Description

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

4

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.

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Problem Setter: Jane Alam Jan

Developed and Maintained by 
JANE ALAM JAN

Copyright © 2012 
LightOJ, Jane Alam Jan

Source

Problem Setter: Jane Alam Jan

在字符串的右边加入元素使之成为回文串---问所能形成回文串的最短长度-.-

设字符串S----反串为T

我们如果直接将T加入到S的后面一定是回文串-.-

让长度尽可能的少-.-就是让T和Ｓ重叠的最多－．－

重叠部分一定是回文串且一定在S的最后面结束----.-

问题就转换为最大的以S末尾结束的回文串的长度-.-

代码“：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char ch[1000100];char T[2000200];int P[2000200];int main(){int t;scanf("%d",&t);for (int ca=1;ca<=t;ca++){scanf("%s",ch);int ll=strlen(ch);memset(T,0,sizeof(T));memset(P,0,sizeof(P));for (int i=0;i<=ll;i++){T[2*i+1]='#';T[2*i+2]=ch[i];}T[0]=-1;ll=2*ll;int mid=0,ss=0,ans=0;for (int i=1;i<=ll;i++){if (ss>i)P[i]=min(ss-i,P[2*mid-i]);elseP[i]=1;while (T[i-P[i]]==T[i+P[i]])P[i]=P[i]+1;if (i+P[i]>ss){ss=i+P[i];mid=i;}if (ss>ll)//以T[i]为中心的字符串以到T的末尾--{ans=P[i]-1;break;}}printf("Case %d: %d\n",ca,ll-ans);}return 0;}