用的 Trie（不是 AC 自动机），TLE 两个点

下面的代码能够拿到除了 #10 #11 之外的全部测试点，正确性肯定有保障，目测复杂度似乎没有问题，可能问题就只有常数太大了，求卡常

#include <bits/stdc++.h>
using namespace std;

bitset<1 << 21> visited;
char s[1 << 21];
int n, m;

class Trie {
	private:
		typedef unsigned int size_T;
		struct Node {
			size_T end_count;
			Node *sons[26];
			Node() {
				end_count = 0;
				memset(sons, 0, sizeof(sons));
			}
		};
		Node *root;
		void dfs(size_T pos, const size_T &length, size_T &answer, const char *word) {
			if (visited[pos] == true)
				return;
			visited[pos] = true;
			answer = pos > answer ? pos : answer;
			Node *current = root;
			for (; pos < length; pos += 1) {
				const size_T &Next = word[pos] - 'a';
				if (current->sons[Next] == 0)
					break;
				current = current->sons[Next];
				if (current->end_count)
					dfs(pos + 1, length, answer, word);
			}
		}
	public:
		Trie() {
			root = new Node;
		}
		void insert(const char *word) {
			Node *current = root;
			for (size_T i = 0; word[i]; i += 1) {
				const size_T &Next = word[i] - 'a';
				if (current->sons[Next] == 0)
					current->sons[Next] = new Node;
				current = current->sons[Next];
			}
			current->end_count += 1;
		}
		size_T get_endCount(const char *word) {
			Node *current = root;
			for (size_T i = 0; word[i]; i += 1) {
				const size_T &Next = word[i] - 'a';
				if (current->sons[Next] == 0)
					return 0;
				current = current->sons[Next];
			}
			return current->end_count;
		}
		size_T query(const char *word) {
			size_T length = strlen(word), answer = 0;
			visited.reset();
			dfs(0, length, answer, word);
			return answer;
		}
} trie;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	cin >> n >> m;
	while (n--) {
		cin >> s;
		trie.insert(s);
	}
	while (m--) {
		cin >> s;
		cout << trie.query(s) << '\n';
	}
	return 0;
}