CF Watto and Mechanism (字典树+深搜)

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


同样的思路，错了很多次，看来实现能力还是不够强。
字典树加搜索，没什么好说的，注意必须要改变一个字母就行。
 1 #include <bits/stdc++.h>
 2 using    namespace    std;
 3 
 4 const    int    SIZE = 800000;
 5 struct    Node
 6 {
 7     Node    * next[3];
 8     bool    word;
 9 }* ROOT = nullptr;
10 char    S[SIZE];
11 
12 bool    dfs(char *,Node *,bool);
13 int    main(void)
14 {
15     int    n,m;
16     char    ch;
17     ROOT = new    Node;
18     for(int i = 0;i < 3;i ++)
19     {
20         ROOT -> next[i] = nullptr;
21         ROOT -> word = false;
22     }
23 
24     cin >> n >> m;
25     cin.ignore();
26     for(int i = 0;i < n;i ++)
27     {
28         Node    * cur = ROOT;
29         while(cin.get(ch) && ch >= 'a' && ch <= 'c')
30         {
31             if(cur -> next[ch - 'a'])
32                 cur = cur -> next[ch - 'a'];
33             else
34             {
35                 cur -> next[ch - 'a'] = new Node;
36                 cur = cur -> next[ch - 'a'];
37                 for(int i = 0;i < 3;i ++)
38                     cur -> next[i] = nullptr;
39                 cur -> word = false;
40             }
41         }
42         cur -> word = true;
43     }
44     for(int i = 0;i < m;i ++)
45     {
46         cin >> S;
47         if(dfs(S,ROOT,true))
48             puts("YES");
49         else
50             puts("NO");
51     }
52 
53     return    0;
54 }
55 
56 bool    dfs(char * s,Node * t,bool chance)
57 {
58     if(!t)
59         return    false;
60     if(!(*s))
61     {
62         if(t -> word && !chance)
63             return    true;
64         return    false;
65     }
66 
67     if(dfs(s + 1,t -> next[*s - 'a'],chance))
68         return    true;
69     if(chance)
70         for(int i = 0;i < 3;i ++)
71             if(i != (*s - 'a'))
72                 if(dfs(s + 1,t -> next[i],false))
73                         return    true;
74     return    false;
75 }

 

posted @ 2015-04-23 14:22  Decouple  阅读(309)  评论(0)  编辑  收藏  举报