hdu 3695 10 福州 现场 F - Computer Virus on Planet Pandora 暴力 ac自动机
Description
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
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
Sample Input
Sample Output
#include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> #include <queue> using namespace std; struct Trie { int next[500010][26],fail[500010],end[500010]; int root,L; int newnode() { for(int i = 0;i < 26;i++) next[L][i] = -1; end[L++] = 0; return L-1; } void init() { L = 0; root = newnode(); } void insert(char buf[]) { int len = strlen(buf); int now = root; for(int i = 0;i < len;i++) { if(next[now][buf[i]-‘A‘] == -1) next[now][buf[i]-‘A‘] = newnode(); now = next[now][buf[i]-‘A‘]; } end[now]++; } void build() { queue<int>Q; fail[root] = root; for(int i = 0;i < 26;i++) if(next[root][i] == -1) next[root][i] = root; else { fail[next[root][i]] = root; Q.push(next[root][i]); } while( !Q.empty() ) { int now = Q.front(); Q.pop(); for(int i = 0;i < 26;i++) if(next[now][i] == -1) next[now][i] = next[fail[now]][i]; else { fail[next[now][i]]=next[fail[now]][i]; Q.push(next[now][i]); } } } int query(char buf[]) { int len = strlen(buf); int now = root; int res = 0; for(int i = 0;i < len;i++) { now = next[now][buf[i]-‘A‘]; int temp = now; while( temp != root ) { res += end[temp]; end[temp] = 0; temp = fail[temp]; } } return res; } void debug() { for(int i = 0;i < L;i++) { printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]); for(int j = 0;j < 26;j++) printf("%2d",next[i][j]); printf("]\n"); } } }; char buf[5100010]; char buf2[5100010]; Trie ac; void rever(char * arr,int len){ len--; for(int i=0;i<=len/2;i++)swap(arr[i],arr[len-i]); } void read(int ind,int & ans,int & gap){ ans=0;gap=0; for(int i=ind;buf[i]<=‘9‘&&buf[i]>=‘0‘;i++){ gap++; ans*=10; ans+=buf[i]-‘0‘; } } int main() { int T; int n; scanf("%d",&T); while( T-- ) { scanf("%d",&n); ac.init(); for(int i = 0;i < n;i++) { scanf("%s",buf); ac.insert(buf); } ac.build(); scanf("%s",buf); int i=0,j=0; for(i=0,j=0;buf[i];){ if(buf[i]<=‘Z‘&&buf[i]>=‘A‘)buf2[j++]=buf[i++]; else if(buf[i]==‘[‘){ int len,gap; read(i+1,len,gap); i+=gap+1; for(int k=0;k<min(len,1005);k++)buf2[j++]=buf[i]; i+=2; } } buf2[j]=0; int ans=ac.query(buf2); rever(buf2,j); ans+=ac.query(buf2); printf("%d\n",ans); } return 0; }
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