Repository

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2538    Accepted Submission(s): 990

Problem Description

When
you go shopping, you can search in repository for avalible merchandises
by the computers and internet. First you give the search system a name
about something, then the system responds with the results. Now you are
given a lot merchandise names in repository and some queries, and
required to simulate the process.

Input

There
is only one case. First there is an integer P
(1<=P<=10000)representing the number of the merchanidse names in
the repository. The next P lines each contain a string (it‘s length
isn‘t beyond 20,and all the letters are lowercase).Then there is an
integer Q(1<=Q<=100000) representing the number of the queries.
The next Q lines each contains a string(the same limitation as foregoing
descriptions) as the searching condition.

Output

For each query, you just output the number of the merchandises, whose names contain the search string as their substrings.

Sample Input

20
ad
ae
af
ag
ah
ai
aj
ak
al
ads
add
ade
adf
adg
adh
adi
adj
adk
adl
aes
5
b
a
d
ad
s

Sample Output

0
20
11
11
2

Source

2009 Multi-University Training Contest 4 - Host by HDU

题意： 给出一些字符，然后寻问一个字符，在给出字符里能找到子串的个数..

代码：

 1 //#define LOCAL
 2 #include<cstdio>
 3 #include<cstring>
 4 typedef struct node
 5 {
 6    struct node *child[26];
 7    int cnt;  //作为统计
 8    int id;
 9 }Trie;
10
11 void Insert(char *s,Trie *root,int id)
12 {
13     int pos,i;
14     Trie *cur=root,*curnew;
15     for(;*s!=‘\0‘;s++)
16     {
17         pos=*s-‘a‘;
18       if(cur->child[pos]==NULL)
19       {
20            curnew = new Trie;
21            for( i=0; i<26;i++)
22             curnew->child[i]=NULL;
23             curnew->cnt=0;
24             curnew->id=0;
25             cur->child[pos]=curnew;
26       }
27       cur=cur->child[pos];
28       if(cur->id!=id) //避免同一个单词重复计算子串
29          {
30              cur->cnt++;
31              cur->id=id;
32          }
33     }
34 }
35
36 int query(char *s, Trie *root)
37 {
38     int pos;
39     Trie *cur=root;
40     while(*s!=‘\0‘)
41     {
42         pos=*s-‘a‘;
43         if(cur->child[pos]==NULL)
44             return 0;
45         cur=cur->child[pos];
46         s++;
47     }
48     return cur->cnt;
49 }
50 void del(Trie *root)
51 {
52     Trie *cur=root;
53     for(int i=0;i<26;i++)
54     {
55         if(cur->child[i]!=NULL)
56            del(cur->child[i]);
57     }
58     delete cur;
59   return ;
60 }
61 char str[22];
62 int main()
63 {
64 #ifdef LOCAL
65   freopen("test.in","r",stdin);
66 #endif
67   int n,m,i;
68   scanf("%d",&n);
69   Trie *root=new Trie;
70   for( i=0;i<26;i++)
71     root->child[i]=NULL;
72     root->cnt=0;
73   while(n--)
74   {
75       scanf("%s",str);
76       for(i=0 ; str[i]!=‘\0‘ ;i++)
77         Insert(str+i,root,n+1);
78   }
79   scanf("%d",&m);
80   while(m--)
81   {
82       scanf("%s",str);
83       printf("%d\n",query(str,root));
84   }
85   del(root);
86   return 0;
87 }