Trie（Codeforces Round #291 (Div. 2)）

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 withn strings. Then the mechanism should be able to process queries of the following type: "Given strings, determine if the memory of the mechanism contains stringt that consists of the same number of characters ass 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 ofn initial lines and m queries. He decided to entrust this job to you.

Input

The first line contains two non-negative numbers n andm (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 consistsonly 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 3aaaaaacacacaaabaaccacacccaaac

Output

YESNONO

思路：构建trie树后，每次判断，对每个位置枚举每种字母替换（因为只有三种字母），然后进行判断

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=1000010;const int SIGMA_SIZE=3;int N,M;char str[maxn];struct Trie{    int ch[maxn][4];    int val[maxn];    int sz;    void clear(){memset(ch[0],0,sizeof(ch[0]));sz=1;}    int idx(char x){return x-'a';}    void insert(char *s,int v)    {        int u=0;        int n=strlen(s);        for(int i=0;i<n;i++)        {            int c=idx(s[i]);            if(!ch[u][c])            {                memset(ch[sz],0,sizeof(ch[sz]));                val[sz]=0;                ch[u][c]=sz++;            }            u=ch[u][c];        }        val[u]=v;    }    bool solve(char *s)    {        int u=0;        int n=strlen(s);        for(int i=0;i<n;i++)        {            int c=idx(s[i]);            for(int j=0;j<SIGMA_SIZE;j++)            {                if(c==j||ch[u][j]==0)continue;                int u2=ch[u][j];                bool flag=true;                for(int k=i+1;k<n;k++)                {                    int c2=idx(s[k]);                    if(!ch[u2][c2])                    {                        flag=false;                        break;                    }                    u2=ch[u2][c2];                }                if(flag&&val[u2])return true;            }            if(!ch[u][c])return false;            u=ch[u][c];        }        return false;    }}tree;int main(){    while(scanf("%d%d",&N,&M)!=EOF)    {        tree.clear();        for(int i=1;i<=N;i++)        {            scanf("%s",str);            tree.insert(str,1);        }        while(M--)        {            scanf("%s",str);            if(tree.solve(str))printf("YES\n");            else printf("NO\n");        }    }    return 0;}