poj3974_Palindrome_ Manacher求最长回文子串的长度

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Palindrome
Time Limit: 15000MS         Memory Limit: 65536K
Total Submissions: 8974         Accepted: 3376
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

abcbabcbabcba
abacacbaaaab
END
Sample Output

Case 1: 13
Case 2: 6

题意：求最长回文串的长度

分析：Manacher的O（n）解法，参考资料：https://www.felix021.com/blog/read.php?2040

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#include<iostream>
#include<cstdio>
#include<string> 
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
 
const int maxn=1000010;
char str[maxn];//原字符串
char tmp[maxn<<1];//转换后的字符串
int Len[maxn<<1];
//转换原始串
int init(char *st)
{
    int i,len=strlen(st);
    tmp[0]='@';//字符串开头增加一个特殊字符，防止越界
    for(i=1;i<=2*len;i+=2)
    {
        tmp[i]='#';
        tmp[i+1]=st[i/2];
    }
    tmp[2*len+1]='#';
    tmp[2*len+2]='$';//字符串结尾加一个字符，防止越界
    tmp[2*len+3]=0;
    return 2*len+1;//返回转换字符串的长度
}
//Manacher算法计算过程
int manacher(char *st,int len)
{
     int mx=0,ans=0,po=0;//mx即为当前计算回文串最右边字符的最大值
     for(int i=1;i<=len;i++)
     {
         if(mx>i)
         Len[i]=min(mx-i,Len[2*po-i]);//在Len[j]和mx-i中取个小
         else
         Len[i]=1;//如果i>=mx，要从头开始匹配
         
         while(st[i-Len[i]]==st[i+Len[i]])
                  Len[i]++;
         if(Len[i]+i>mx)//若新计算的回文串右端点位置大于mx，要更新po和mx的值
         {
             mx=Len[i]+i;
             po=i;
         }
         ans=max(ans,Len[i]);
     }
     return ans-1;//返回Len[i]中的最大值-1即为原串的最长回文子串额长度
}
int main()
{
    int r=1;
    while(~scanf("%s",str))
    {
        if(str[0]=='E') break;
        int l=init(str);
        printf("Case %d: ",r++);
        cout<<manacher(tmp,l)<<endl;
    }
}