HDU 3695 computer Virus on Planet Pandora

Computer Virus on Planet PandoraTime limit:6000/2000 MS (java/others) Memory limit:256000/128000 K (java/others)
Total submission (s): 2480 Accepted Submission (s): 688


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 has anti-virus software. Of course they learned virus scanning algorithm from the Earth. Every virus has a pattern string which consists of the capital letters. If A virus ' s pattern string is a substring of a program, or the pattern string was a substring of the reverse of that progr AM, they can say the program was 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.

Inputthere is multiple test cases. The first line of the input was an integer T (t<=) indicating the number of test cases.

For each test case:

The first line is a integer n (0 < n <=) 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 is all different so there
is n different viruses. The length of pattern string is no more than $ and a pattern string at least consists of one letter.

The last line of a test was the program. The program is 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 are a capital letter. It means q consecutive letter Xs on the original uncompressed program. For example, [6K] means
' Kkkkkk ' in the original program. So, if a compressed program was like:

Ab[2d]e[7k]g

It actually is abddekkkkkkkg through decompressed to original format.

The length of the program was at least 1 and at the very 5,100,000, no matter in the compressed format or under it is decompres SED to original format.

Outputfor each of the test case, print an integer K in a line meaning that the program is infected by K viruses.

Sample Input

32abdcbdacb3abccdeghiabccdefihg4abbacdeebbbfeeea[2b]cd[4e]f

Sample Output

032Hintin the "second case" in the sample input, the reverse of the program was ' GHIFEDCCBA ', and ' GHI ' is a Substri Ng of the reverse, so the program was infected by virus ' GHI '.

Source2010 Asia Fuzhou Regional Contest problem-solving ideas: AC automata naked problem, to find the positive and negative two strings of the total number of words appear

#include <iostream> #include <cstdio> #include <cstring> #include <queue> #define N 5555555using    Namespace std;struct node{node *fail;    Node *next[26];    int cnt;        Node () {fail=0;        cnt=0;    memset (Next,0,sizeof (next));    }};node *root=null;void Insert (char *str) {int i=0,index;    Node *p=root;        while (str[i]!= ') {index=str[i]-' A ';        if (p->next[index]==null) p->next[index]=new node;        p=p->next[index];    i++; } p->cnt++;}    void build () {int i;    root->fail=null;    Queue<node*> Q;    Q.push (root);        while (!q.empty ()) {node *tmp=q.front ();        Q.pop ();        Node *p=null;                    for (i=0;i<26;i++) {if (Tmp->next[i]) {if (tmp==root)                tmp->next[i]->fail=root;                    else {p=tmp->fail; while (p) {                        if (P->next[i]) {tmp->next[i]->fail=p-&                            Gt;next[i];                        Break                    } p=p->fail;                } if (P==null) tmp->next[i]->fail=root;            } q.push (Tmp->next[i]);    }}}}int query (char *str) {int i=0,cnt=0,index;    Node *p=root;        while (str[i]!= ') {//cout<<i<<endl;        index=str[i]-' A ';        while (P->next[index]==null&&p!=root) p=p->fail;        p=p->next[index];        p= (p==null) root:p;        Node *tmp=p;            while (tmp!=root&&tmp->cnt!=-1) {cnt+=tmp->cnt;            tmp->cnt=-1;        tmp=tmp->fail;    } i++; } return CNT;}    Char s1[n],s2[n];void input () {int p=0;    char c;    GetChar (); while (C=getchar (),C!= ' \ n ') {if (c!= ' [') s1[p++]=c;            else {int x;            scanf ("%d", &x);            C=getchar ();            while (x--) s1[p++]=c;        GetChar ();    }} for (int q=0;p>0;q++) s2[q]=s1[--p]; cout<<s1<< "" <<s2<<endl;}    void del (node *root) {for (int i=0;i<26;i++) if (Root->next[i]) del (root->next[i]); Delete root;}    int main () {//Freopen ("In.txt", "R", stdin);//Freopen ("OUT.txt", "w", stdout);    int t,n;    scanf ("%d", &t);    Char str[1005];        while (t--) {root=new node;        memset (S1, ' n ', sizeof (S1));        memset (S2, ' n ', sizeof (S2));        scanf ("%d", &n);        while (n--) scanf ("%s", str), insert (str);        Build ();        Input ();        printf ("%d\n", Query (S1) +query (S2));    del (root); } return 0;}