URAL1960-Palindromes and Super Abilities

最新推荐文章于 2019-09-12 18:30:23 发布

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本文介绍了一个利用回文树解决回文串计数问题的方法，具体为在线性时间内计算给定字符串所有前缀中不同回文串的数量。通过构建回文树的数据结构，实现对输入字符串的逐字符添加并更新回文串种类数量。

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Palindromes and Super Abilities

Time limit: 1.0 second
Memory limit: 64 MB

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 s1, ..., sn 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 s1...sn, where si are small English letters (1 ≤ n ≤ 105).

Output

Output n numbers separated by whitespaces, i-th of these numbers must be the number of different nonempty substrings of prefix s1...si that are palindromes.

Sample

input	output
aba	1 2 3

Problem Author: Mikhail Rubinchik (prepared by Grigory Nazarov)
Problem Source: Ural FU contest. Kontur Cup. Petrozavodsk training camp. Winter 2013

题意：给你一个字符串，求出每个前缀有多少种回文串

解题思路：回文树，回文树每开一个新节点就是一种新的回文串

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const int maxn = 3e5 + 10;
char s[maxn];

struct PalindromicTree
{
	const static int maxn = 3e5 + 10;
	int next[maxn][26], last, sz, tot;
	int fail[maxn], len[maxn];
	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 Node(int length)
	{
		memset(next[sz], 0, sizeof(next[sz]));
		len[sz] = length;
		return sz++;
	}
	int getfail(int x)
	{
		while (s[tot] != s[tot - len[x] - 1]) x = fail[x];
		return x;
	}
	int add(char pos)
	{
		int x = (s[++tot] = pos) - 'a', y = getfail(last);
		if (next[y][x]) { last = next[y][x]; return 0; }
		last = next[y][x] = Node(len[y] + 2);
		fail[last] = len[last] == 1 ? 2 : next[getfail(fail[y])][x];
		return 1;
	}
}solve;

int main()
{
	while (~scanf("%s", s))
	{
		solve.Clear();
		int ans = 0, flag = 0;
		for (int i = 0; s[i]; i++)
		{
			ans += solve.add(s[i]);
			if (flag) printf(" ");
			else flag = 1;
			printf("%d", ans);
		}
		printf("\n");
	}
	return 0;
}