HDOJ4287 Intelligent IME[字典树水题]

Intelligent IME

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 145 Accepted Submission(s): 80

Problem Description

　　We all use cell phone today. And we must be familiar with the intelligent English input method on the cell phone. To be specific, the number buttons may correspond to some English letters respectively, as shown below:
　　2 : a, b, c 3 : d, e, f 4 : g, h, i 5 : j, k, l 6 : m, n, o
　　7 : p, q, r, s 8 : t, u, v 9 : w, x, y, z
　　When we want to input the word “wing”, we press the button 9, 4, 6, 4, then the input method will choose from an embedded dictionary, all words matching the input number sequence, such as “wing”, “whoi”, “zhog”. Here comes our question, given a dictionary, how many words in it match some input number sequences?

Input

　　First is an integer T, indicating the number of test cases. Then T block follows, each of which is formatted like this:
　　Two integer N (1 <= N <= 5000), M (1 <= M <= 5000), indicating the number of input number sequences and the number of words in the dictionary, respectively. Then comes N lines, each line contains a number sequence, consisting of no more than 6 digits. Then comes M lines, each line contains a letter string, consisting of no more than 6 lower letters. It is guaranteed that there are neither duplicated number sequences nor duplicated words.

Output

　　For each input block, output N integers, indicating how many words in the dictionary match the corresponding number sequence, each integer per line.

Sample Input

1 3 5 46 64448 74 go in night might gn

Sample Output

3 2 0

Source

2012 ACM/ICPC Asia Regional Tianjin Online

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liuyiding

题意:下面的字符串中，有多少个是按照相应的数字输入顺序得到的。

code:

 1 #include<iostream>
 2 using namespace std;
 3 
 4 int n,m;
 5 char str[5100][8];
 6 int ans;
 7 
 8 struct node
 9 {
10     int cnt;
11     node *next[26];
12 }*p;
13 
14 int key[26]={2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9};
15 
16 void build(char str[],int k,node *head)
17 {
18     while(k<strlen(str))
19     {
20         if(head->next[str[k]-'a']!=NULL)
21         {
22             head->next[str[k]-'a']->cnt+=1;
23             head=head->next[str[k]-'a'];
24         }
25         else
26         {
27             head->next[str[k]-'a']=new node;
28             head=head->next[str[k]-'a'];
29             head->cnt=1;
30             for(int i=0;i<26;i++)
31                 head->next[i]=NULL;
32         }
33         k++;
34     }
35 }
36 
37 void search(char str[],int k,node *head)
38 {
39     if(head==NULL)
40         return;
41     if(k==strlen(str))
42     {
43         ans+=head->cnt;
44         return;
45     }
46     for(int i=0;i<26;i++)
47     {
48         if(key[i]-(str[k]-'0')==0)
49         {
50             search(str,k+1,head->next[i]);
51         }
52     }
53 }
54 
55 int main()
56 {
57     int t;
58     scanf("%d",&t);
59     while(t--)
60     {
61         p=new node;
62         for(int i=0;i<26;i++)
63             p->next[i]=NULL;
64         scanf("%d%d",&n,&m);
65         for(int i=0;i<n;i++)
66             scanf("%s",str[i]);
67         for(int i=0;i<m;i++)
68         {
69             char strt[20];
70             scanf("%s",strt);
71             build(strt,0,p);
72         }
73         for(int i=0;i<n;i++)
74         {
75             ans=0;
76             search(str[i],0,p);
77             printf("%d\n",ans);
78         }
79     }
80     return 0;
81 }