CF291-C

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


给定n个字符串,然后再给m次询问,每次询问包含一个字符串
问当前询问的字符串是否和之前n个字符串中某个字符串只有一个位置的字符不相同
有就输出YES,否则输出NO
很恶心的一题,赛中和赛后WA了20+,只能怪自己智商不够
暴力能过,但是要分类来暴力才能过,想不到
常规方法是:trie树+dfs,正好能练下trie树,就敲了
必须得使用dfs来暴力遍历trie树,这里之前因为没用dfs,wa了10+
dfs过程记录不符字符个数,到根时看是否为1即可

#include <iostream>
using namespace std;
#include <string.h>
char str[600100];
int ok[600100];
int len;
int count;
bool flag;
class TrieTree
{
public:
    TrieTree *next[3];//要初始化！！否则指针会乱指!
    bool isleaf;
    TrieTree()
    {
        isleaf=false;
        for(int i=0;i<3;i++)
            next[i]=NULL;
    }
};
void BuildTree(TrieTree *T,int i)
{
    if(T->next[str[i]-'a']==NULL)
        T->next[str[i]-'a']=new TrieTree();
    if(i==len-1)
    {
        T->next[str[i]-'a']->isleaf=true;
        return;
    }
    BuildTree(T->next[str[i]-'a'],i+1);
}
void dfs(TrieTree *T,int i)
{
    if(flag||count>1)
        return;
    if(i==len)
    {
        if(count==1)
        {
            flag=true;
            cout<<"YES"<<endl;
        }
        return;
    }
    for(int j=0;j<3;j++)
        if(T->next[j]!=NULL)
        {
            if(j!=str[i]-'a')
            {
                count++;
                dfs(T->next[j],i+1);
                count--;
            }
            else
                dfs(T->next[j],i+1);
        }
}
int main()
{
    int n,m;
    cin>>n>>m;
    TrieTree T;
    for(int i=1;i<=n;i++)
    {
        cin>>str;
        len=strlen(str);
        ok[len]=1;
        BuildTree(&T,0);
    }
    for(int i=1;i<=m;i++)
    {
        cin>>str;
        len=strlen(str);
        if(!ok[len])
        {
            cout<<"NO"<<endl;
            continue;
        }
        count=0;
        flag=false;
        dfs(&T,0);
        if(!flag)
            cout<<"NO"<<endl;
    }
    return 0;
}