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.
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).
For each test case in the input print the test case number and the length of the largest palindrome.
#include <iostream>
#include <cstdio>
#include <cstring>
#define REP(i, s, n) for(int i = s; i <= n; i ++)
#define REP_(i, s, n) for(int i = n; i >= s; i --)]
#define max(a, b) a > b ? a : b
#define min(a, b) a < b ? a : b
#define MAX_N 100000 + 10
using namespace std;
char s[(MAX_N << 1) + 1], ts[MAX_N];
int len[(MAX_N << 1) + 1], id = 0;
int main(){
while(scanf("%s", ts + 1) != EOF){
if(ts[1] == ‘E‘) break;
memset(len, 0, sizeof(len));
int l = strlen(ts + 1);
REP(i, 1, l) s[i * 2 - 1] = ts[i];
REP(i, 0, l) s[i * 2] = ‘#‘;
l = 2 * l;
int ans = 0, mx = 0, pos = 0;
REP(i, 0, l){
if(mx > i) len[i] = min(len[(pos << 1) - i], mx - i);
else len[i] = 1;
while(i - len[i] >= 0 && i + len[i] <= l && s[i + len[i]] == s[i - len[i]]) len[i]++;
if(mx < i + len[i]) mx = len[i] + i, pos = i;
ans = max(ans, len[i]);
}
printf("Case %d: %d\n", ++ id, ans - 1);
}
return 0;
}
