Computer Virus on Planet Pandora
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 256000/128000 K (Java/Others) Total Submission(s): 1846 Accepted Submission(s): 510
Problem Description
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 ( T<= 10) indicating the number of test cases.
For each test case:
The first line is a integer n( 0 < n <= 250) 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
For each test case:
The first line is a integer n( 0 < n <= 250) 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”.
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 < q <= 5,000,000)and x is a capital letter. It means q consecutive letter xs in the original uncompressed program. For example, [6K] means ‘KKKKKK’
[qx]
q is a number( 0 < q <= 5,000,000)and x is a capital letter. It means q consecutive letter xs 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.
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.
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
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’.模板题：
1 #pragma comment(linker, "/STACK:16777216") 2 #include <iostream> 3 #include <stdio.h> 4 #include <stdlib.h> 5 #include <algorithm> 6 #include <math.h> 7 #include <queue> 8 #include <string.h> 9 using namespace std; 10 11 typedef struct abcd 12 { 13 abcd *next[26]; 14 int end; 15 abcd *fail; 16 }abcd; 17 int ans; 18 abcd *inti() 19 { 20 abcd *t; 21 t=(abcd *)malloc(sizeof(abcd)); 22 for(int i=0;i<26;i++) 23 t->next[i]=NULL; 24 t->end=0; 25 t->fail=NULL; 26 return t; 27 } 28 void insert(abcd *t,char z[]) 29 { 30 if(*z==' ') 31 { 32 t->end++; 33 return; 34 } 35 if(t->next[*z-'A']==NULL) 36 t->next[*z-'A']=inti(); 37 insert(t->next[*z-'A'],z+1); 38 } 39 void ac(abcd *t) 40 { 41 queue<abcd*>a; 42 while(!a.empty())a.pop(); 43 for(int i=0;i<26;i++) 44 { 45 if(t->next[i]!=NULL) 46 { 47 t->next[i]->fail=t; 48 a.push(t->next[i]); 49 } 50 else t->next[i]=t; 51 } 52 abcd *r,*f; 53 while(!a.empty()) 54 { 55 r=a.front(); 56 a.pop(); 57 for(int i=0;i<26;i++) 58 { 59 if(r->next[i]) 60 { 61 a.push(r->next[i]); 62 f=r->fail; 63 while(!f->next[i])f=f->fail; 64 r->next[i]->fail=f->next[i]; 65 } 66 } 67 } 68 } 69 void query(abcd *t,char x[]) 70 { 71 abcd *f,*p=t; 72 while(*x) 73 { 74 while(!p->next[*x-'A'])p=p->fail; 75 p=p->next[*x-'A']; 76 f=p; 77 while(f!=t&&f->end!=-1) 78 { 79 ans+=f->end; 80 f->end=-1; 81 f=f->fail; 82 } 83 x++; 84 } 85 x--; 86 p=t; 87 while(*x) 88 { 89 while(!p->next[*x-'A'])p=p->fail; 90 p=p->next[*x-'A']; 91 f=p; 92 while(f!=t&&f->end!=-1) 93 { 94 ans+=f->end; 95 f->end=-1; 96 f=f->fail; 97 } 98 x--; 99 } 100 } 101 void del(abcd *t) 102 { 103 for(int i=0;i<26;i++) 104 { 105 if(!t->next[i]) 106 del(t->next[i]); 107 } 108 free(t); 109 } 110 int main() 111 { 112 int tr,n,i; 113 //cout<<"YES"<<endl; 114 cin>>tr; 115 while(tr--) 116 { 117 abcd *t; 118 t=inti(); 119 char z[1009]; 120 cin>>n; 121 for(i=0;i<n;i++) 122 { 123 cin>>z; 124 insert(t,z); 125 } 126 ac(t); 127 char x[5200000]; 128 int tt=1,r; 129 char xx; 130 x[0]=' '; 131 xx=getchar(); 132 xx=getchar(); 133 while(xx!=' ') 134 { 135 if(xx!='[') 136 x[tt++]=xx; 137 else 138 { 139 cin>>r; 140 xx=getchar(); 141 for(int i=0;i<r;i++) 142 x[tt++]=xx; 143 xx=getchar(); 144 } 145 xx=getchar(); 146 } 147 x[tt]=' '; 148 ans=0; 149 query(t,x+1); 150 cout<<ans<<endl; 151 del(t); 152 } 153 }