Codeforces Round #291 (Div. 2) -- C. Watto and Mechanism

C. Watto and Mechanism

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Watto, the owner of a spare parts store, has recently got an order for the mechanism that can process strings in a certain way. Initially the memory of the mechanism is filled with n strings. Then the mechanism should be able to process queries of the following type: "Given string s, determine if the memory of the mechanism contains string t that consists of the same number of characters as s and differs froms in exactly one position".

Watto has already compiled the mechanism, all that's left is to write a program for it and check it on the data consisting of n initial lines and m queries. He decided to entrust this job to you.

Input

The first line contains two non-negative numbers n and m (0 ≤ n ≤ 3·105, 0 ≤ m ≤ 3·105) — the number of the initial strings and the number of queries, respectively.

Next follow n non-empty strings that are uploaded to the memory of the mechanism.

Next follow m non-empty strings that are the queries to the mechanism.

The total length of lines in the input doesn't exceed 6·105. Each line consists only of letters 'a', 'b', 'c'.

Output

For each query print on a single line "YES" (without the quotes), if the memory of the mechanism contains the required string, otherwise print "NO" (without the quotes).

Sample test(s)

input

2 3
aaaaa
acacaca
aabaa
ccacacc
caaac

output

YES
NO
NO

思路：刚开始我还以为是Trie树，，然后慢慢的敲。。。最后发现暴力就可以了，，而且貌似很多人用hash做得，，不太会，，以后再慢慢来try吧。。这题先暴力解决掉

暴力的AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <set>
#define LL long long
using namespace std;

int main() {
	int n, m;
	while(scanf("%d %d", &n, &m) != EOF) {
		set<string> a;
		LL maxn = 0;
		for(int i = 0; i < n; i++) {
			string tmp;
			cin >> tmp;
			a.insert(tmp);
			if((int)tmp.size() > maxn) maxn = tmp.size();
		}
		
		if(m * n * maxn < 100000000) {
			for(int i = 0 ; i < m; i++) {
				string tmp;
				cin >> tmp;
				bool flag = false;
				for(set<string>::iterator it = a.begin(); it != a.end() && !flag; it++) {
					if(it->size() == tmp.size()) {
						int cnt = 0;
						for(int j = 0; j < it->size(); j++) {
							if((*it)[j] != tmp[j]) cnt++;
						}
						if(cnt == 1) flag = true;
					}
				}
				if(flag) printf("YES\n");
				else printf("NO\n");
			}
			continue;
		}
		
		for(int i = 0; i < m; i++) {
			string tmp;
			cin >> tmp;
			bool flag = false;
			for(int j = 0; j < tmp.size() && !flag; j++) {
				if(tmp[j] == 'a') {
					tmp[j] = 'b';
					if(a.find(tmp) != a.end()) flag = true;
					tmp[j] = 'c';
					if(a.find(tmp) != a.end()) flag = true;
					tmp[j] = 'a';
				}
				else if(tmp[j] == 'b') {
					tmp[j] = 'c';
					if(a.find(tmp) != a.end()) flag = true;
					tmp[j] = 'a';
					if(a.find(tmp) != a.end()) flag = true;
					tmp[j] = 'b';
				}
				else if(tmp[j] == 'c') {
					tmp[j] = 'a';
					if(a.find(tmp) != a.end()) flag = true;
					tmp[j] = 'b';
					if(a.find(tmp) != a.end()) flag = true;
					tmp[j] = 'c';
				}
			}
			if(flag) printf("YES\n");
			else printf("NO\n");
		}
	}
	return 0;
}