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;

}