URAL 1960 Palindromes and Super Abilities

Description

After solving seven problems on Timus Online Judge with a word “palindrome” in the problem name, Misha has got an unusual ability. Now, when he reads a word, he can mentally count the number of unique nonempty substrings of this word that are palindromes.

Dima wants to test Misha’s new ability. He adds letters  s ₁, ...,  s _n to a word, letter by letter, and after every letter asks Misha, how many different nonempty palindromes current word contains as substrings. Which  n numbers will Misha say, if he will never be wrong?

Input

The only line of input contains the string  s ₁...  s _n, where  s _i are small English letters (1 ≤  n ≤ 10  ⁵).

Output

Output  n numbers separated by whitespaces,  i-th of these numbers must be the number of different nonempty substrings of prefix  s ₁...  s _ithat are palindromes.

Sample Input

input	output
aba	1 2 3

回文树模板题，练练手

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int T, n;
char s[maxn];

struct PalindromicTree
{
	const static int maxn = 1e5 + 10;
	const static int size = 26;
	int next[maxn][size], last;
	int fail[maxn], len[maxn], sz, tot;
	char s[maxn];
	void clear() 
	{ 
		len[1] = -1; len[2] = 0;
		fail[2] = fail[1] = 1;	
		last = (sz = 3) - 1;	tot = 0;
		memset(next[1], 0, sizeof(next[1]));
		memset(next[2], 0, sizeof(next[2]));
	}
	int add(char pos)
	{
		int x = (s[++tot] = pos) - 'a';
		int y = last, z = len[last];
		while (true)
		{
			if (tot - 1 - z > 0 && pos == s[tot - 1 - z]) break;
			y = fail[y];	z = len[y];
		}
		if (next[y][x]) { last = next[y][x]; return 0; }

		last = next[y][x] = sz;
		len[sz] = len[y] + 2;
		memset(next[sz], 0, sizeof(next[sz]));
		if (len[sz] == 1) { fail[sz++] = 2;	return 1; }

		while (true)
		{
			y = fail[y];	z = len[y];
			if (tot - 1 - z > 0 && pos == s[tot - 1 - z]) break;
		}
		fail[sz++] = next[y][x];
		return 1;
	}
}solve;

int main()
{
	while (scanf("%s", s) != EOF)
	{
		solve.clear();
		int ans = 0;
		for (int i = 0; s[i]; i++)
		{
			ans += solve.add(s[i]);
			printf("%s%d", i ? " " : "", ans);
		}
		putchar(10);
	}
	return 0;
}