Palindrome（Manacher求最大回文长度）

Link：http://poj.org/problem?id=3974

Palindrome

Time Limit: 15000MS Memory Limit: 65536KTotal Submissions: 5719 Accepted: 2073

Description

Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient algorithm to find the length of the largest palindrome in a string?" 

A string is said to be a palindrome if it reads the same both forwards and backwards, for example "madam" is a palindrome while "acm" is not. 

The students recognized that this is a classical problem but couldn't come up with a solution better than iterating over all substrings and checking whether they are palindrome or not, obviously this algorithm is not efficient at all, after a while Andy raised his hand and said "Okay, I've a better algorithm" and before he starts to explain his idea he stopped for a moment and then said "Well, I've an even better algorithm!". 

If you think you know Andy's final solution then prove it! Given a string of at most 1000000 characters find and print the length of the largest palindrome inside this string.

Input

Your program will be tested on at most 30 test cases, each test case is given as a string of at most 1000000 lowercase characters on a line by itself. The input is terminated by a line that starts with the string "END" (quotes for clarity). 

Output

For each test case in the input print the test case number and the length of the largest palindrome. 

Sample Input

abcbabcbabcbaabacacbaaaabEND

Sample Output

Case 1: 13Case 2: 6

Source

Seventh ACM Egyptian National Programming Contest

AC code：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>#include<map>#define LL long long#define MAXN 1000010using namespace std;char s1[MAXN*2],s2[MAXN*2];//s1为原字符串，s2为转换后的字符串int p[MAXN*2];//p[i]表示以s2[i]为中心(包括s2[i])的回文半径,对应s1的回文长度 char ans[MAXN*2];//原字符串的最长回文串 int len; int l,r;//最长回文子串对应原串T中的位置：l = (i - p[i])/2; r = (i + p[i])/2 - 2;void pre(){int i,j,k;len=strlen(s1);s2[0]='$';s2[1]='#';for(i=0;i<len;i++){s2[i*2+2]=s1[i];s2[i*2+3]='#';}len=len*2+2;s2[len]=0;}int Manacher(){int i;int mx=0;int id;int ans=0;for(i=1;i<len;i++){if(mx>i)p[i]=min(p[2*id-i],mx-i);elsep[i]=1;for(;s2[i+p[i]]==s2[i-p[i]];p[i]++);if(ans<p[i]){ans=p[i];l = (i - p[i])/2;//最长回文子串对应原串s1中的左位置 r = (i + p[i])/2 - 2;//最长回文子串对应原串s1中的右位置}//ans=max(ans,p[i]);if(p[i]+i>mx){mx=p[i]+i;id=i;}}return ans-1;//返回最长回文子串长度 }int main(){int cas=0;while(~scanf("%s",s1)){if(strcmp(s1,"END")==0)break; cas++;pre();printf("Case %d: %d\n",cas,Manacher());}return 0;}