POJ 3974 Palindrome（Manacher）

Palindrome

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

题目大意：给出一个字符串，求它的最长回文子串的长度。
解题思路：Manacher算法的模板题……Manacher算法可以在O(N)时间内求出以各个位置为中心的最长回文子串的长度。（但是这个算法还没有完全搞懂，晚上回去再研究研究）

代码如下：

#include <algorithm>
#include <cctype>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#define EPS 1e-6
#define INF INT_MAX / 10
#define LL long long
#define MOD 100000000
#define PI acos(-1.0)

const int maxn = 2000005;

char s[maxn];
int p[maxn];
int lens;
int manacher()
{
int mx,id,ans;
mx = 0,id = 0,ans = 0;
memset(p,0,sizeof(p));
for(int i = 1;s[i] != '\0';i++){
p[i] = mx > i ? std::min(p[2 * id - i], mx - i) : 1;
while(s[i + p[i]] == s[i - p[i]])
p[i]++;
if (i + p[i] > mx){
mx = i + p[i];
id = i;
}
ans = std::max(ans,p[i]);
}
return ans - 1;
}

int main()
{
char ch;
int ncase = 0;
while(1){
int i = 1,j = 2;
while((ch = getchar()) != '\n'){
s[i] = '#';
s[j] = ch;
i += 2;
j += 2;
}
lens = strlen(s + 1);
s[++lens] = '#';
s[++lens] = '\0';
s[0] = '$';
if(strcmp(s + 1,"#E#N#D#") == 0)
break;
printf("Case %d: %d\n",++ncase,manacher());
memset(s,0,sizeof(s));
}
return 0;
}