Keywords Search
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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
Sample Output
AC 自動(dòng)機(jī)
我的代碼：
1 #include <iostream>
2 #include <cstdio>
3
4 using namespace std;
5
6 const int ACTC = 26;
7 const int ACTM = 800000;
8 const int ACQL = 800000;
9
10 struct AC
11 {
12 int count;
13 AC * fail;
14 AC * ch[ ACTC ];
15 };
16
17 AC * que[ ACQL ];
18
19 AC * ac_new( bool init = false ) {
20 int i;
21 AC * p;
22 static AC memAC[ ACTM ];
23 static int tot = 0;
24 if ( init ) {
25 tot = 0;
26 return 0;
27 }
28 p = memAC + tot++;
29 p->count = 0;
30 p->fail = 0;
31 for ( i = 0; i < ACTC; ++i )
32 p->ch[ i ] = 0;
33 return p;
34 }
35
36 int ac_add( AC * & root, const char * first, const char * last ) {
37 AC ** p = &root;
38 for ( ; ; ) {
39 if ( *p == 0 ) *p = ac_new();
40 if ( first == last ) return ++( (*p)->count );
41 p = &( (*p)->ch[ *first++ ] );
42 }
43 }
44
45 void ac_build( AC * root ) {
46 // root != 0
47 int qh = 0, qt = 1, i;
48 AC *pf, *pc, *pt;
49 root->fail = 0;
50 que[ 0 ] = root;
51 while ( qh != qt ) {
52 pf = que[ qh ];
53 qh = ( qh + 1 ) % ACQL;
54 for ( i = 0; i < ACTC; ++i ) {
55 if ( pc = pf->ch[ i ] ) {
56 for ( pt = pf->fail; pt && ( pt->ch[ i ] == 0 ); pt = pt->fail )
57 ;
58 pc->fail = pt ? pt->ch[ i ] : root;
59 que[ qt ] = pc;
60 qt = ( qt + 1 ) % ACQL;
61 }
62 }
63 }
64 }
65
66 int ac_query( AC * root, const char * first, const char * last ) {
67 // root != 0
68 int ans = 0;
69 AC *p = root, *q;
70 while ( first != last ) {
71 while ( p && ( p->ch[ *first ] == 0 ) ) {
72 p = p->fail;
73 }
74 if ( p ) {
75 q = p = p->ch[ *first++ ];
76 while ( q && ( q->count != -1 ) ) {
77 ans += q->count;
78 q->count = -1;
79 q = q->fail;
80 }
81 }
82 else {
83 p = root;
84 ++first;
85 }
86 }
87 return ans;
88 }
89
90 char txt[ 1000009 ], pat[ 70 ];
91 AC * root;
92
93 int main() {
94 int td, n;
95 char * pc;
96 scanf( "%d", &td );
97 while ( td-- ) {
98 scanf( "%d%*c", &n );
99 ac_new( true );
100 root = 0;
101 while ( n-- ) {
102 gets( pat );
103 for ( pc = pat; *pc; ++pc )
104 *pc -= 'a';
105 ac_add( root, pat, pc );
106 }
107 gets( txt );
108 for ( pc = txt; *pc; ++pc )
109 *pc -= 'a';
110 ac_build( root );
111 printf( "%d\n", ac_query( root, txt, pc ) );
112 }
113 return 0;
114 }
115