Computer Virus on Planet Pandora Description， ACM/ICPC 2010/2011 亚洲，福州区域赛 F， UVA 5103
| 5103 - Computer Virus on Planet Pandora Description Asia - Fuzhou - 2010/2011 |
Aliens on planet Pandora also write computer programs like us. Their programs only consist of capital letters (` A' to ` Z') which they learned from the Earth. On planet Pandora, hackers make computer virus, so they also have anti-virus software. Of course they learned virus scanning algorithm from the Earth. Every virus has a pattern string which consists of only capital letters. If a virus's pattern string is a substring of a program, or the pattern string is a substring of the reverse of that program, they can say the program is infected by that virus. Give you a program and a list of virus pattern strings, please write a program to figure out how many viruses the program is infected by.
Input
There are multiple test cases. The first line in the input is an integer T (T10) indicating the number of test cases.
For each test case:
The first line is a integer n (0 < n250) indicating the number of virus pattern strings.
Then n lines follows, each represents a virus pattern string. Every pattern string stands for a virus. It's guaranteed that those n pattern strings are all different so there are n different viruses. The length of pattern string is no more than 1,000 and a pattern string at least consists of one letter.
The last line of a test case is the program. The program may be described in a compressed format. A compressed program consists of capital letters and ``compressors". A ``compressor" is in the following format:
[qx]
q is a number (0 < q5, 000, 000) and x is a capital letter. It means q consecutive letter x's in the original uncompressed program. For example, [6K] means ` KKKKKK' in the original program. So, if a compressed program is like:
AB[2D]E[7K]G
it actually is ABDDEKKKKKKKG after decompressed to original format.
The length of the program is at least 1 and at most 5,100,000, no matter in the compressed format or after it is decompressed to original format.
Output
For each test case, print an integer K in a line meaning that the program is infected by K viruses.
Hint: In the second case in the sample input, the reverse of the program is ` GHIFEDCCBA', and ` GHI' is a substring of the reverse, so the program is infected by virus ` GHI'.
Sample Input
3
2
AB
DCB
DACB
3
ABC
CDE
GHI
ABCCDEFIHG
4
ABB
ACDEE
BBB
FEEE
A[2B]CD[4E]F
Sample Output
0
3
2
Fuzhou 2010-2011
AC 自动机。。。
2 #include < cstdio >
3
4 using namespace std;
5
6 const int ACTC = 26 ;
7 const int ACTM = 200000 ;
8 const int ACQL = ACTM;
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[ 5100009 ], pat[ 1009 ];
91 AC * root;
92
93 void expand( char * out ) {
94 char ch;
95 char * q = out ;
96 ch = getchar();
97 while ( (ch != ' \n ' ) && (ch != EOF) ) {
98 if ( ( ' A ' <= (ch)) && ((ch) <= ' Z ' ) ) {
99 * q ++ = ch;
100 }
101 else {
102 ch = getchar();
103 int num = 0 ;
104 while ( ( ' 0 ' <= (ch)) && ((ch) <= ' 9 ' ) ) {
105 num = num * 10 + ( ch - ' 0 ' );
106 ch = getchar();
107 }
108 while ( num -- > 0 ) {
109 * q ++ = ch;
110 }
111 ch = getchar();
112 }
113 ch = getchar();
114 }
115 * q = 0 ;
116 }
117
118 int main() {
119 int td, n, ans;
120 char * pc, * ph, * pt;
121 scanf( " %d " , & td );
122 while ( td -- ) {
123 scanf( " %d%*c " , & n );
124 ac_new( true );
125 root = 0 ;
126 while ( n -- ) {
127 gets( pat );
128 for ( pc = pat; * pc; ++ pc )
129 * pc -= ' A ' ;
130 ac_add( root, pat, pc );
131 }
132 ac_build( root );
133 expand( txt );
134 for ( pc = txt; * pc; ++ pc )
135 * pc -= ' A ' ;
136 ans = ac_query( root, txt, pc );
137 ph = txt;
138 pt = pc - 1 ;
139 while ( ph < pt ) {
140 char tmp = * ph;
141 * ph = * pt;
142 * pt = tmp;
143 ++ ph;
144 -- pt;
145 }
146 ans += ac_query( root, txt, pc );
147 printf( " %d\n " , ans );
148 }
149 return 0 ;
150 }
151