HDU 2222 Keywords Search

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 57724 Accepted Submission(s): 18954

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

Author

Wiskey

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Statistic | Submit | Discuss | Note

给出一些A串，和一个B串，询问有多少个A串是B的子串。

先对所有A串构造Trie树，然后建成AC自动机。这样用B串在自动机中做匹配时，每每匹配到一个结点t，在这个结点以及t->fail,t->fail->fail...等结点上结束的A串都是在B串中匹配上的。因此应当从t一直顺fail指针走到root，同时注意不要重复统计相同的A串。

 1 #include <bits/stdc++.h>
 2 
 3 struct node {
 4     int count;
 5     node *fail;
 6     node *next[26];
 7     node(void) {
 8         memset(this, 0, sizeof(node));
 9     }
10 }*root;
11 
12 inline void insert(char *s) {
13     node *t = root;
14     for ( ; *s; ++s) {
15         if (t->next[*s - 'a'] == NULL)
16             t->next[*s - 'a'] = new node;
17         t = t->next[*s - 'a'];
18     }
19     ++t->count;
20 }
21 
22 inline void build(void) {
23     root->fail = root;
24     std::queue<node *> q;
25     for (int i = 0; i < 26; ++i)
26         if (root->next[i]) {
27             root->next[i]->fail = root;
28             q.push(root->next[i]);
29         }
30     while (!q.empty()) {
31         node *t = q.front(); q.pop();
32         for (int i = 0; i < 26; ++i)
33             if (t->next[i]) {
34                 node *p = t->fail;
35                 while (p != root && !p->next[i])
36                     p = p->fail;
37                 if (p->next[i])
38                     t->next[i]->fail = p->next[i];
39                 else
40                     t->next[i]->fail = root;
41                 q.push(t->next[i]);
42             }
43     }
44 }
45 
46 inline int count(char *s) {
47     int ans = 0;
48     node *t = root;
49     for ( ; *s; ++s) {
50         while (t != root && !t->next[*s - 'a'])
51             t = t->fail;
52         if (t->next[*s - 'a'])
53             t = t->next[*s - 'a'];
54         for (node *p = t; p != root && p->count != -1; p = p->fail)
55             ans += p->count, p->count = -1;
56     }
57     return ans;
58 }
59 
60 signed main(void) {
61     int cas, n; scanf("%d", &cas);
62     for (char s[1000005]; cas--; ) {
63         root = new node;
64         scanf("%d", &n);
65         while (n--) {
66             scanf("%s", s);
67             insert(s);
68         }
69         build();
70         scanf("%s", s);
71         printf("%d
", count(s));
72     }
73 }

@Author: YouSiki