Keywords Search
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10744 Accepted Submission(s): 3722
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.
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.
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
该题为AC自动机模板题，果然它还是很强大的，该题要注意单词有重复出现的。
代码如下：
1 #include <cstring>
2 #include <cstdio>
3 #include <cstdlib>
4 #include <queue>
5 using namespace std;
6
7 struct Node
8 {
9 int end;
10 Node *fail, *child[26];
11 };
12
13 char s[1000005];
14
15 void getstr( char *s )
16 {
17 char c;
18 int p = 0;
19 while( c = getchar(), c == ' ' || c == '\n' ) ;
20 s[p++] = c;
21 while( c = getchar(), c != ' ' && c != '\n' )
22 s[p++] = c;
23 s[p] = '\0';
24 }
25
26 Node *init( )
27 {
28 Node *r = ( Node * )malloc( sizeof( Node ) );
29 r->end = 0;
30 r->fail = NULL;
31 memset( r->child, NULL, sizeof( r->child ) );
32 return r;
33 }
34
35 void creattree( Node *p, char *in )
36 {
37 int index = *in - 'a';
38 if( p->child[index] == NULL )
39 p->child[index] = init();
40 if( *( in + 1 ) )
41 creattree( p->child[index], in + 1 );
42 else
43 p->child[index]->end++;
44 }
45
46 void creatfail( Node *p )
47 {
48 queue< Node * >q;
49 q.push( p );
50 while( !q.empty() )
51 {
52 Node *pos = q.front();
53 q.pop();
54 for( int i = 0; i < 26; ++i )
55 {
56 Node *f = pos->fail;
57 if( pos->child[i] )
58 {
59 while( f != NULL )
60 {
61 if( f->child[i] )
62 {
63 pos->child[i]->fail = f->child[i];
64 break;
65 }
66 else
67 f = f->fail;
68 }
69 if( f == NULL )
70 {
71 pos->child[i]->fail = p;
72 }
73 q.push( pos->child[i] );
74 }
75 }
76 }
77 }
78
79 int acauto( Node *p, char *text )
80 {
81 int ans = 0, lenth = 0, len = strlen( text );
82 Node *f = p;
83 for( int i = 0; i < len; ++i )
84 {
85 int type = text[i] - 'a';
86 while( f )
87 {
88 if( f->child[type] )
89 {
90 f = f->child[type];
91 break;
92 }
93 else
94 f = f->fail;
95 }
96
97 if( f == NULL ) // 如果都已经找到了根节点仍不匹配
98 {
99 f = p; // 从头开始
100 continue;
101 }
102
103 Node *t = f;
104 while( t != p ) // 成功匹配一个字符的更新操作
105 {
106 if( t->end )
107 {
108 ans += t->end;
109 t->end = 0;
110 }
111 t = t->fail;
112 }
113 }
114 return ans;
115 }
116
117 void _free( Node *p )
118 {
119 for( int i = 0; i < 26; ++i )
120 {
121 if( p->child[i] )
122 _free( p->child[i] );
123 }
124 free( p );
125 }
126
127 int main()
128 {
129 int T;
130 scanf( "%d\n", &T );
131 while( T-- )
132 {
133 int N;
134 Node *p = init();
135 scanf( "%d", &N );
136 for( int i = 0; i < N; ++i )
137 {
138 getstr( s );
139 creattree( p, s );
140 }
141 getstr( s );
142 creatfail( p );
143 printf( "%d\n", acauto( p, s ) );
144 }
145 }