Computer Virus on Planet PandoraTime Limit: 6000/2000 MS (Java/Others)Memory Limit: 256000/128000 K (Java/Others)Total Submission(s): 2847Accepted Submission(s): 799
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. Onplanet 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 thereare 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]GIt 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
32ABDCBDACB3ABCCDEGHIABCCDEFIHG4ABBACDEEBBBFEEEA[2B]CD[4E]F
Sample Output
032
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
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AC自动机水题
/*************************************************************************> File Name: hdu3695.cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: 2015年02月04日 星期三 21时17分18秒 ************************************************************************/#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const double pi = acos(-1);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;const int N = 5110000;char str[1010];char buf[N];char input[N];struct node{node *next[26]; //根据题目而定node *fail;int count;node (){fail = NULL;for (int i = 0; i < 26; ++i){next[i] = NULL;}count = 0;}};node *qu[500010];class AC_Automation{private:node *root;int head;int tail;public:AC_Automation(){head = 0;tail = 0;root = new node();}void Build_Trie(char str[]){int len = strlen (str);node *p = root;for (int i = 0; i < len; ++i){int ind = str[i] – 'A';if (p -> next[ind] == NULL){p -> next[ind] = new node();}p = p -> next[ind];}++p -> count;}void Build_AC (){root -> fail = NULL;qu[tail++] = root;while (head < tail){node *now = qu[head++];node *p = NULL;for (int i = 0; i < 26; ++i){if (now -> next[i] != NULL){if (now == root){now -> next[i] -> fail = root;}else{p = now -> fail;while (p != NULL){if (p -> next[i] != NULL){now -> next[i] -> fail = p -> next[i];break;}else{p = p -> fail;}}}if (p == NULL){now -> next[i] -> fail = root;}qu[tail++] = now -> next[i];}}}}int Match (char buf[]){int ans = 0;int len = strlen (buf);node *p = root;for (int i = 0; i < len; ++i){int ind = buf[i] – 'A';while (p -> next[ind] == NULL && p != root){p = p -> fail;}p = p -> next[ind];if (p == NULL){p = root;}node *tmp = p;while (tmp != root && tmp -> count != -1){ans += tmp -> count;tmp -> count = -1;tmp = tmp -> fail;}}return ans;}void dfs (node *p){for (int i = 0; i < 26; ++i){if (p -> next[i] != NULL){dfs (p -> next[i]);}}delete p;}~AC_Automation(){dfs (root);}};int main (){int t;scanf("%d", &t);while (t–){int n;scanf("%d", &n);AC_Automation AC;for (int i = 1; i <= n; ++i){scanf("%s", str);AC.Build_Trie (str);}scanf("%s", input);int len = strlen (input);int cnt = 0;for (int i = 0; i < len; ++i){if (input[i] >= 'A' && input[i] <= 'Z'){buf[cnt++] = input[i];}else if (input[i] == '['){int x = 0;++i;while (input[i] >= '0' && input[i] <= '9' && i < len){x = x * 10 + input[i++] – '0';}for (int j = 0; j < x; ++j){buf[cnt++] = input[i];}++i;}else if (input[i] == ']'){continue;}}buf[cnt] = '\0';AC.Build_AC();int ans = AC.Match(buf);for (int i = 0;i < cnt / 2; ++i){swap (buf[i], buf[cnt – i – 1]);}ans += AC.Match(buf);printf("%d\n", ans);}return 0;}
，不要害怕错过什么，因为在路上你就已经收获了自由自在的好心情。
