Palindrome（manacher算法）

Palindrome

Time Limit: 15000MS		Memory Limit: 65536K
Total Submissions: 11080		Accepted: 4212

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

Source

Seventh ACM Egyptian National Programming Contest

题意：寻找 给出字符串 中最长的回文字串，输出它的长度；

思路：manacher算法-----利用最短时间，找出最长回文字串的长度；

manacher算法详解参考博客：https://blog.youkuaiyun.com/dyx404514/article/details/42061017

                                              https://segmentfault.com/a/1190000008484167#articleHeader2

此题代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
const int mm=1000005;
char a[mm],tmp[mm*2];
int p[mm*2];
int init()
{
    int la=strlen(a);
    tmp[0]='$';                         //最左端分界符
    tmp[1]='#';                         //字母之间的分隔符
    int j=2;
    for(int i=0;i<la;i++)
    {
        tmp[j++]=a[i];             //转换字符串
        tmp[j++]='#';
    }
    tmp[j]='\0';
    return j;
}
int mach()
{
    int len=init();     
    int id,mx=0,maxx=-1;             //id是最大回文字串中心，mx是回文字串能延伸到最右端的位置。
    for(int i=1;i<len;i++)
    {  
        if(mx>i)
            p[i]=min(p[2*id-i],mx-i);          
        else
            p[i]=1;                            //i>=mx 
        while(tmp[i-p[i]]==tmp[i+p[i]])     
            p[i]++;
        if(mx<p[i]+i)
        {
             id=i;                          //更新id，mx的值；
             mx=p[i]+i;          
        }
        maxx=max(maxx,p[i]-1);    
    }
    return maxx;
}
int main()
{
    int ca=1;   
    while(~scanf("%s",a))
    {
        if(!strcmp(a,"END"))
            break;   
        else
            printf("Case %d: %d\n",ca++,mach());                
         mem(p,0);
    }
    return 0;
}