URAL2040-Palindromes and Super Abilities 2

本文介绍了一种使用回文树解决字符串问题的方法,具体为按顺序增加字符时判断新增加的字符是否引入新的回文子串。通过构建回文树并更新节点状态,该算法能在O(n)时间内高效完成任务。

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Palindromes and Super Abilities 2

Time limit: 0.5 second
Memory limit: 100 MB
Dima adds letters s1, …, sn one by one to the end of a word. After each letter, he asks Misha to tell him how many new palindrome substrings appeared when he added that letter. Two substrings are considered distinct if they are different as strings. Which n numbers will be said by Misha if it is known that he is never wrong?

Input

The input contains a string s1 … sn consisting of letters ‘a’ and ‘b’ (1 ≤ n ≤ 5 000 000).

Output

Print n numbers without spaces: i-th number must be the number of palindrome substrings of the prefix s1 … si minus the number of palindrome substrings of the prefix s1 … si−1. The first number in the output should be one.

Sample

input output
abbbba
111111

Notes

We guarantee that jury has C++ solution which fits Time Limit at least two times. We do not guarantee that solution on other languages exists (even Java).
Problem Author: Mikhail Rubinchik (prepared by Kirill Borozdin)
Problem Source: Ural FU Dandelion contest. Petrozavodsk training camp. Summer 2014


题意:给你一个字符串,按字符串的顺序每次增加一个字符的时候,如果增加了回文串种类,那就输出1,否则输出0

解题思路:回文树,但是会卡输出


#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 = 5e6 + 10;
char s[maxn], ans[maxn];

struct PalindromicTree
{
	const static int maxn = 5e6 + 10;
	int next[maxn][2], 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 len = strlen(s);
		for (int i = 0; i < len; i++) ans[i] = '0' + solve.add(s[i]);
                ans[len] = '\0';
		printf("%s\n", ans);
	}
	return 0;
}
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