C. Watto and Mechanism 字典树 Codeforces Round #291 (Div. 二)

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

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

题意：给n个字符串和m次询问，每次询问的字符串如果能够由前面n个字符串中的某一个只改变一个字母得到输出YES，否则NO。

用字典树解决，渣渣不熟悉字典树，写在这里以后多看看。。。。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 601005
#define MAXN 3
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
typedef long long ll;
using namespace std;

typedef struct TrieNode
{
    int nCount; //标记该结点是否是某个字符串的结尾
    struct TrieNode *next[MAXN];
}TrieNode;

char s[maxn];
int n,m;
TrieNode *proot;

TrieNode *CreateTrieNode() 
{
    TrieNode *p;
    p=(TrieNode *)malloc(sizeof(TrieNode));
    p->nCount=0;
    for (int i=0;i<3;i++)
        p->next[i]=NULL;
    return p;
}

void Trie_insert(char *s)  //建字典树
{
    TrieNode *p;
    p=proot;
    if (!p)
        p=proot=CreateTrieNode();
    int i=0;
    while (s[i])
    {
        int k=s[i++]-'a';
        if (!p->next[k])
            p->next[k]=CreateTrieNode();
        p=p->next[k];
    }
    p->nCount=1;
}

bool dfs(TrieNode *root,int len,int flag)  //dfs搜索
{
    if (s[len])
    {
        int k=s[len]-'a';
        if (root->next[k]!=NULL)
        {
            if (dfs(root->next[k],len+1,flag))
                return true;
        }
        if (!flag)
        {
            for (int i=0;i<3;i++)
                if (i!=k&&root->next[i]!=NULL)
                    if (dfs(root->next[i],len+1,++flag))
                        return true;
        }
    }
    else
    {
        if (flag&&root->nCount==1)
            return true;
    }
    return false;
}

int main()
{
    while (~scanf("%d%d",&n,&m))
    {
        proot=NULL;
        for (int i=0;i<n;i++)
        {
            scanf("%s",s);
            Trie_insert(s);
        }
        for (int i=0;i<m;i++)
        {
            scanf("%s",s);
            if (dfs(proot,0,0)) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}

C. Watto and Mechanism 字典树 Codeforces Round #291 (Div. 二)

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