AC自动机 HDU2222 Keywords Search

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 62674    Accepted Submission(s): 20749


Problem Description
In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.
 

 

Input
First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.
 

 

Output
Print how many keywords are contained in the description.
 

 

Sample Input
1 5 she he say shr her yasherhs
 

 

Sample Output
3
 

 

Author
Wiskey
 
 
模板题，模板源自chaijing
 
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 using namespace std;
 7 int t,n,num;
 8 int len,val,j;
 9 char s[1000010];
10 struct data{
11     int ch[30];
12     int fail,cnt;//cnt 以该节点结束的词的个数 
13 }tree[1000010];
14 queue<int>q;
15 void clear(int x){
16     memset(tree[x].ch,0,sizeof(tree[x].ch));
17     tree[x].fail=0;
18     tree[x].cnt=0;
19 }
20 void trie(int x){
21     len=strlen(s);
22     for(int i=0;i<len;i++){
23         val=s[i]-'a'+1;
24         if(!tree[x].ch[val]){
25             num++;
26             clear(num);
27             tree[x].ch[val]=num;
28         }
29         x=tree[x].ch[val];
30     }
31     tree[x].cnt++;
32 }
33 void build(){
34     while(!q.empty()) q.pop();
35     for(int i=1;i<=26;i++) 
36         if(tree[0].ch[i]) q.push(tree[0].ch[i]);
37     while(!q.empty()){
38         int p=q.front();
39         q.pop();
40         for(int i=1;i<=26;i++){
41             int fail=tree[p].fail;
42             if(tree[p].ch[i]){
43                 
44                 tree[tree[p].ch[i]].fail=tree[fail].ch[i];
45                 q.push(tree[p].ch[i]);
46             }
47             else tree[p].ch[i]=tree[fail].ch[i];
48         }
49     }
50 }
51 int find(int x){
52     int ans=0;
53     len=strlen(s);
54     for(int i=0;i<len;i++){
55         val=s[i]-'a'+1;
56         while(x&&!tree[x].ch[val]) x=tree[x].fail;
57         x=tree[x].ch[val];
58         j=x;
59         while(j&&tree[j].cnt!=-1){
60             ans+=tree[j].cnt;
61             tree[j].cnt=-1;
62             j=tree[j].fail;
63         }
64     }
65     return ans;
66 }
67 int main(){
68     scanf("%d",&t);
69     while(t){
70         t--;
71         num=0;
72         clear(0);
73         scanf("%d",&n);
74         for(int i=1;i<=n;i++){
75             scanf("%s",s);
76             trie(0);
77         }
78         build();
79         scanf("%s",s);
80         printf("%d\n",find(0));
81     }
82     return 0;
83 }

 

 
 
posted @ 2017-05-29 00:13  zwube  阅读(127)  评论(0)  编辑  收藏  举报