POJ3974 Palindrome(Manacher模板题)

 POJ3974 Palindrome

描述

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.InputYour 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).OutputFor each test case in the input print the test case number and the length o
f the largest palindrome.
给出一个长度为N(1 <= N <= 1000000)的字符串，求最长回文子串的长度。

样例

输入

abcbabcbabcba
abacacbaaaab
END

输出

Case 1: 13
Case 2: 6

#include<bits/stdc++.h>
using namespace std;
char s[1000001],s_new[3000000];
int p[3000000];
int get()
{
    s_new[0]='$';
    s_new[1]='#';
    int k=2;
    int len=strlen(s);
    for(int i=0;i<len;i++)
    {
        s_new[k++]=s[i];
        s_new[k++]='#';
    }
    s_new[k]='';
    return k;
}
int manacher()
{
    int len=get();
    int id,mx=0;
    int minn=-1;
    for(int i=1;i<=len;i++)
    {
        if(i<mx)
        p[i]=min(mx-i,p[2*id-i]);
        else
        p[i]=1;
        while(s_new[i-p[i]]==s_new[i+p[i]])
        p[i]++;
        if(mx<i+p[i])
        {
            id=i;
            mx=i+p[i];
        }
        minn=max(minn,p[i]-1);
    }
    return minn;
}
int main()
{
    int k=1;
    while(1)
    {
        cin>>s;
        if(s[0]=='E'&&s[1]=='N'&&s[2]=='D')
        break;
        cout<<"Case "<<k<<": "<<manacher()<<endl;
        k++;
    }
    return 0;
}