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 from s 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

题意：输入n+m串字符串，判断m中的每个字符串是否在前n个字符串中有满足条件（1.长度相同;2.最多只有一个字符不一样）的对应字符串；
分析：字典树+dfs  据说暴力也能过；

#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;

const int MAXN=1000010; //其实300000以上就够 习惯开大些
int dic[MAXN][5],val[MAXN],cnt,flag,ok;

void inst(char *s)　　//插入Trie树 数组为链
{
    int u=0,v,len;
    len=strlen(s);
    for(int i=0;i<len;i++)
    {
        v=s[i]-'a'+1; 　　//分别标记 a b c 为 1 2 3  其实0 1 2 也可以 不过为了与初始化的值区分开 便于查错 就这样做了
        if(!dic[u][v]) 　　//如果此节点未被编码过  
        {
            dic[u][v]=++cnt;　　//编码此数组 其实就是形成链表的过程
            memset(dic[cnt],0,sizeof(dic[cnt]));　　//初始化该节点下所属的字母
        }
        u=dic[u][v]; 　　//取出该节点编码 作用：1、便于对该节点赋值操作 2、生成链表或继续访问链表
        if(i==len-1) val[u]=1;　　//如果此时该字符串已结束 标记结束时的编码值
    }
}

bool dfs(char  *s,int r,int u,int num) //括号里面的内容分别为  字符串s 字符下标r 字符的节点u 不同字母个数num
{
    if(s[r]) //如果该字符串没有结束
    {
        int v=s[r]-'a'+1;　　// a b c 转换为 1 2 3 与插入时保持一致 
        if(dic[u][v]) 　　//如果该字母与树上的字母有匹配的连接方式
        {
            if(dfs(s,r+1,dic[u][v],num)) {return true;}　　//继续深搜 一直到底 如果中途不满足条件就会跳出
        }
        if(!num) 　　//num最多有1个 只有为0的时候才可替换此处字母继续访问链标 
        {
            for(int i=1;i<=3;i++)
            {
                if(i!=v&&dic[u][i])　　 //与原来字母不相同 并且树上存在匹配方式
                    if(dfs(s,r+1,dic[u][i],num+1)) 　　//相当与把当前字母替换为其他字母 然后继续访问
                    return true;
            }
        }
    }
    else if(num&&val[u]) return true;　　//字符串已结束 判断是否有同样长度的字符串
    return false;
}

int main()
{
    int n,m;
    char s[MAXN];　　 //这里如果用string会暴MEMORY  这里不大明白为什么 谅

    memset(dic[0],0,sizeof(dic[0]));　　//一次性全部初始化也可
    memset(val,0,sizeof(val));    //初始化所有编码值
    cnt=0;　　　　//初始化编码号

    scanf("%d%d",&n,&m);
    getchar();
    for(int i=0;i<n;i++)　　//插入字符操作 一开始考虑用set去存字符串 结果查看测试数据时发现其实没必要
    {
        scanf("%s",s);
        inst(s);
    }

    for(int i=0;i<m;i++) 
    {
        scanf("%s",&s);
        flag=dfs(s,0,0,0); 　　//flag标记该串是否满足条件
        if(flag) printf("YES
");
        else printf("NO
");
    }
}