# Watto and Mechanism - CodeForces 514 C Trie树 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 ≤ 3·1050 ≤ 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).

AC代码如下：

``````#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
int T,t,n,m;
char s;
struct Trie
{
int index;
Trie *next;
Trie()
{
index=-1;
memset(next,0,sizeof(next));
}
};
Trie *root=new Trie;
void Trie_Insert(Trie *tr,int len)
{
if(s[len]!='\0')
{
if(tr->next[s[len]-'a']==0)
tr->next[s[len]-'a']=new Trie;
Trie_Insert(tr->next[s[len]-'a'],len+1);
}
else
tr->index=1;
}
bool dfs(Trie *tr,int len,int f)
{
if(s[len]=='\0' )
{
if(f==0 && tr->index==1)
return true;
else
return false;
}
if(tr->next[s[len]-'a']!=0)
{
if(dfs(tr->next[s[len]-'a'],len+1,f))
return true;
}
if(f==1)
{
int i;
for(i=0;i<=2;i++)
if(i!=s[len]-'a' && tr->next[i]!=0)
if(dfs(tr->next[i],len+1,0))
return true;
}
return false;
}
int main()
{
int i,j,k;
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
{
scanf("%s",s+1);
Trie_Insert(root,1);
}
for(i=1;i<=m;i++)
{
scanf("%s",s+1);
if(dfs(root,1,1))
printf("YES\n");
else
printf("NO\n");
}
}
``````

原文作者：Trie树
原文地址: https://blog.csdn.net/u014733623/article/details/43830229
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