hdu ----3695 Computer Virus on Planet Pandora (ac自动机)
hdu ----3695 Computer Virus on Planet Pandora (ac自动机)
Computer Virus on Planet Pandora
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 256000/128000 K (Java/Others) Total Submission(s): 2975 Accepted Submission(s): 833
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 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 < 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.
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’.
Source
2010 Asia Fuzhou Regional Contest
题目分析： 给定一些特征码，一个目标串，然后要你求包含的特征码的值，值得注意的是，对于目标串需要正反两次扫一遍，然后就是解析目标串，将[23c]解压成正常的字符串形式就可以了。
代码：
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<queue>
4 #include<iostream>
5
6 using namespace std;
7
8 const int maxn=26;
9 struct Trie{
10
11 Trie *child[maxn];
12 Trie * fail ;
13 int tail;
14
15 //函数不占用内存空间
16 void init(){
17 for(int i=0;i<maxn ;i++ )
18 child[i]=NULL;
19 fail=NULL;
20 tail=0;
21 }
22
23 };
24
25 char var[1010] ,stt[5100010];
26 char stmp[5100010]={'\0'};
27
28 //构建一个Trie树
29 void BuildTrie(char st[] , Trie * root){
30
31 Trie * cur ;
32 for( int i=0; st[i] ; i++ ){
33 int ps=st[i]-'A';
34 if(root->child[ps]==NULL){
35 cur = new Trie;
36 cur->init(); //初始化
37 root->child[ps]=cur;
38 }
39 root = root->child[ps];
40 }
41
42 root->tail++;
43 }
44
45 //构造失败指正
46 void BuildFail(Trie * root){
47 queue<Trie *> sav;
48 Trie * tmp,*cur ;
49 sav.push(root);
50 while(!sav.empty()){
51 tmp = sav.front();
52 sav.pop();
53 for( int i=0 ; i<maxn ; i++ ){
54 if(tmp->child[i]!=NULL){
55 if(tmp==root) {
56 tmp->child[i]->fail=root;
57 }
58 else{
59 cur =tmp;
60 while(cur->fail){
61 if(cur->fail->child[i]!=NULL){
62 tmp->child[i]->fail = cur->fail->child[i];
63 break;
64 }
65 cur =cur->fail;
66 }
67
68 if(cur->fail==NULL)
69 tmp->child[i]->fail = root ;
70
71 }
72 sav.push(tmp->child[i]);
73 }
74 }
75 }
76 }
77
78 //查询
79 int Query( char ss[] , Trie *root){
80
81 Trie * tmp=root ,*cur;
82 int res=0,ps=0;
83 for(int i=0; ss[i] ;i++){
84 ps = ss[i]-'A';
85 while(tmp->child[ps]==NULL&&tmp!=root){
86 tmp=tmp->fail;
87 }
88 tmp = tmp->child[ps];
89 if(tmp==NULL) tmp=root;
90 cur=tmp;
91 while(cur!=root&&cur->tail>0){
92 res+=cur->tail;
93 cur->tail=0;
94 cur= cur->fail;
95 }
96 }
97 return res ;
98 }
99
100 //对字符串进行必要的伸展
101 int change(char stt[] ,char stmp []){
102
103 int i,j;
104 for(i=0,j=0; stt[i] ;i++){
105 if(stt[i]!='['){
106 stmp[j++]=stt[i];
107 }else{
108 int k=0,lvar=0;
109 while(stt[++i]>='0'&&stt[i]<='9'){
110 lvar=lvar*10+(stt[i]-'0');
111 }
112 for(k=0;k<lvar;k++){
113 stmp[j++]=stt[i];
114 }
115 i++;
116 }
117 }
118 stmp[j]='\0';
119 return j;
120 }
121
122 //交换字符
123 void swap(char * a,char *b){
124 if(*a!=*b) *a^=*b^=*a^=*b;
125 }
126 //对字符串进行反序输入
127
128 void Reverse(char * ss,int len){
129
130 for(int i=0;i<len/2;i++){
131 swap(ss[i],ss[len-i-1]);
132 }
133 }
134
135 int main(){
136
137 int tes,nm,res;
138 scanf("%d",&tes);
139 while(tes--){
140 res=0;
141 Trie * root = new Trie;
142 root->init();
143 scanf("%d",&nm);
144 while(nm--){
145 scanf("%s",var);
146 BuildTrie(var,root);
147 }
148 BuildFail(root);
149 scanf("%s",stt);
150 int len= change(stt,stmp);
151 res=Query(stmp,root);
152 Reverse(stmp,len);
153 res+=Query(stmp,root);
154 printf("%d\n",res);
155 }
156 return 0;
157 }