poj3974 Palindrome

Palindrome

Time Limit: 15000MS		Memory Limit: 65536K
Total Submissions: 10526		Accepted: 4001

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

大致题意：求最长回文子串的长度.

分析：manacher算法模板.为了避免对奇偶性的讨论，在每个字母的两边加字符#,这样统计#的回文半径就能得到偶数回文子串的长度.回文长度=新的回文半径-1.

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 1e6 + 10;

char s[maxn], ss[maxn * 3];
int cas, r[maxn * 3], n, mx, p, ans;

void solve()
{
    for (int i = 2; i < n; i++)
    {
        int t;
        if (i < mx)
            t = min(mx - i, r[2 * p - i]);
        else
            t = 1;
        while (ss[i + t] == ss[i - t])
            t++;
        r[i] = t;
        if (i + t > mx)
        {
            mx = i + t;
            p = i;
        }
    }
}

int main()
{
    while (scanf("%s", s + 1) != EOF && s[1] != 'E')
    {
        ans = 0;
        p = mx = 1;
        n = 0;
        memset(r, 0, sizeof(r));
        int len = strlen(s + 1);
        ss[++n] = '(';
        for (int i = 1; i <= len; i++)
        {
            ss[++n] = '#';
            ss[++n] = s[i];
        }
        ss[++n] = '#';
        ss[++n] = ')';
        solve();
        for (int i = 1; i <= n; i++)
            ans = max(ans, r[i]);
        ans--;
        printf("Case %d: %d\n", ++cas, ans);
    }

    return 0;
}