传送门: 3430 Detect the Virus                                                                                                                                                                                                                                                                            Detect the Virus


Time Limit: 2 Seconds      Memory Limit: 65536 KB

One day, Nobita found that his computer is extremely slow. After several hours' work, he finally found that it was a virus that made his poor computer slow and the virus was activated by a misoperation of opening an attachment of an email.

Nobita did use an outstanding anti-virus software, however, for some strange reason, this software did not check email attachments. Now Nobita decide to detect viruses in emails by himself.

To detect an virus, a virus sample (several binary bytes) is needed. If these binary bytes can be found in the email attachment (binary data), then the attachment contains the virus.

Note that attachments (binary data) in emails are usually encoded in base64. To encode a binary stream in base64, first write the binary stream into bits. Then take 6 bits from the stream in turn, encode these 6 bits into a base64 character according the following table:

That is, translate every 3 bytes into 4 base64 characters. If the original binary stream contains 3k + 1 bytes, where k is an integer, fill last bits using zero when encoding and append '==' as padding. If the original binary stream contains 3k + 2 bytes, fill last bits using zero when encoding and append '=' as padding. No padding is needed when the original binary stream contains 3k bytes.

Value 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
Encoding A B C D E F G H I J K L M N O P Q R S T U V W X Y Z a b c d e f
Value 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63
Encoding g h i j k l m n o p q r s t u v w x y z 0 1 2 3 4 5 6 7 8 9 + /

For example, to encode 'hello' into base64, first write 'hello' as binary bits, that is: 01101000 01100101 01101100 01101100 01101111
Then, take 6 bits in turn and fill last bits as zero as padding (zero padding bits are marked in bold):
011010 000110 010101 101100 011011 000110 111100
They are
26 6 21 44 27 6 60
in decimal. Look up the table above and use corresponding characters:
aGVsbG8
Since original binary data contains 1 * 3 + 2 bytes, padding is needed, append '=' and 'hello' is finally encoded in base64:
aGVsbG8=

Section 5.2 of RFC 1521 describes how to encode a binary stream in base64 much more detailedly:

Click here to see Section 5.2 of RFC 1521 if you have interest

Here is a piece of ANSI C code that can encode binary data in base64. It contains a function, encode (infile, outfile), to encode binary file infile in base64 and output result to outfile.

Click here to see the reference C code if you have interest

Input

Input contains multiple cases (about 15, of which most are small ones). The first line of each case contains an integer N (0 <= N <= 512). In the next N
distinct lines, each line contains a sample of a kind of virus, which
is not empty, has not more than 64 bytes in binary and is encoded in
base64. Then, the next line contains an integer M (1 <= M <= 128). In the following M
lines, each line contains the content of a file to be detected, which
is not empty, has no more than 2048 bytes in binary and is encoded in
base64.

There is a blank line after each case.

Output

For each case, output M lines. The ith line contains the number of kinds of virus detected in the ith file.

Output a blank line after each case.

Sample Input

3
YmFzZTY0
dmlydXM=
dDog
1
dGVzdDogdmlydXMu 1
QA==
2
QA==
ICAgICAgICA=

Sample Output

2

1
0

Hint

In the first sample case, there are three virus samples: base64, virus and t: , the data to be checked is test: virus., which contains the second and the third, two virus samples.


Author: WU, Jun

Contest: ZOJ Monthly, November 2010

-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Solution: A-C Automation

注意:

1.弄清题意,“If these binary bytes can be found in the email attachment (binary data), then the attachment contains the virus.“题意是,将base64编码的模式串与文本串翻译成ASCII编码的字符串,再进行匹配。并不是将两者翻译成二进制(0-1串)进行匹配

2.也可将模式串与文本串翻译成0-1串做匹配,但注意匹配的位置必须是整字节处。这样做代码更简洁,内存与时间都更优(Maybe, 因为字符集是{0, 1})。

 #include<bits/stdc++.h>
using namespace std;
const char *m1="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
int m2[];
void build_map(){
for(int i=; m1[i]; i++){
m2[m1[i]]=i;
}
}
const int MAX_N=1e6+, ls=, lt=2e4, MAX_M=;
bool s[ls], t[lt], used[MAX_M];
char v[ls], f[lt];
int q[MAX_N], head, tail;
struct node
{
int pre, id, pos[], last;
void init(){
id=;
memset(pos, , sizeof(pos));
}
};
node trie[MAX_N];
void b2b(char ch, bool *b, int &i){
if(ch=='=') i-=;
else{
int p=;
while(p){
b[i++]=m2[ch]&p;
p>>=;
}
}
}
int trans(char *s, bool *b){
int i=, j=;
for(; s[i]; i++){
b2b(s[i], b, j);
}
//for(int i=0; i<j; i++) printf("%d", b[i]);
//puts("");
return j;
}
int build_trie(int N){
int tot=, now, len;
trie[tot].init();
for(int id=; id<=N; id++){
scanf("%s", v);
len=trans(v, s);
now=;
for(int i=; i<len; i++){
int &nt=trie[now].pos[s[i]]; //error-prone
now=nt?nt:(trie[++tot].init(), nt=tot);
}
trie[now].id=id;
}
return tot;
}
void build_ac(){
head=tail=;
trie[].last=;
for(int i=; i<; i++){
int &nt=trie[].pos[i];
if(nt){
trie[nt].pre=;
trie[nt].last=;
q[tail++]=nt;
}
}
int pre;
while(head!=tail){
int &top=q[head++];
for(int i=; i<; i++){
int &nt=trie[top].pos[i];
pre=trie[top].pre;
if(!nt) nt=trie[pre].pos[i];
else{
q[tail++]=nt;
while(!trie[pre].pos[i]&&pre) pre=trie[pre].pre;
pre=trie[nt].pre=trie[pre].pos[i];
int &last=trie[nt].last;
last=trie[pre].id?pre:trie[pre].last; //error-prone
}
}
}
}
void get_ans(int pre, int &cnt){ //error-prone
pre=trie[pre].id?pre:trie[pre].last;
while(pre && !used[trie[pre].id]){
used[trie[pre].id]=true;
cnt++;
pre=trie[pre].last;
}
}
void match(int N){
while(N--){
scanf("%s", f);
int len=trans(f, t);
memset(used, , sizeof(used));
int cnt=;
for(int i=, pre=; i<len; i++){
pre=trie[pre].pos[t[i]];
if(i%==){
get_ans(pre, cnt);
}
}
printf("%d\n", cnt);
}
} int main(){
//freopen("in", "r", stdin);
build_map();
int N, M;
while(~scanf("%d", &N)){
build_trie(N);
build_ac();
scanf("%d", &M);
match(M);
puts("");
}
return ;
}

ZOJ 3430 Detect the Virus的更多相关文章

  1. zoj 3430 Detect the Virus(AC自己主动机)

    题目连接:zoj 3430 Detect the Virus 题目大意:给定一个编码完的串,将每个字符相应着表的数值转换成6位二进制.然后以8为一个数值,又一次形成字符 串,推断给定询问串是否含有字符 ...

  2. ZOJ - 3430 Detect the Virus —— AC自动机、解码

    题目链接:https://vjudge.net/problem/ZOJ-3430 Detect the Virus Time Limit: 2 Seconds      Memory Limit: 6 ...

  3. zoj 3430 Detect the Virus(AC自己主动机)

    Detect the Virus Time Limit: 2 Seconds      Memory Limit: 65536 KB One day, Nobita found that his co ...

  4. ZOJ 3430 Detect the Virus(AC自动机)

    题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3430 题意:给你n个编码后的模式串,和m个编码后的主串,求原来主 ...

  5. ZOJ 3430 Detect the Virus 【AC自动机+解码】

    解码的那些事儿,不多说. 注意解码后的结果各种情况都有,用整数数组存储,char数组会超char类型的范围(这个事最蛋疼的啊)建立自动机的时候不能用0来判断结束. #include <cstdi ...

  6. ZOJ 3430 Detect the Virus(AC自动机 + 模拟)题解

    题意:问你主串有几种模式串.但是所有串都是加密的,先解码.解码过程为:先把串按照他给的映射表变成6位数二进制数,然后首尾衔接变成二进制长串,再8位8位取变成新的数,不够的补0.因为最多可能到255,所 ...

  7. ZOJ 4114 Detect the Virus(AC自动机)

    Detect the Virus Time Limit: 2 Seconds      Memory Limit: 65536 KB One day, Nobita found that his co ...

  8. 【ZOJ】3430 Detect the Virus

    动态建树MLE.模仿别人的代码模板各种原因wa后,终于AC. #include <iostream> #include <cstdio> #include <cstrin ...

  9. Detect the Virus ZOJ - 3430 AC自动机

    One day, Nobita found that his computer is extremely slow. After several hours' work, he finally fou ...

随机推荐

  1. [转]VMware虚拟机上网络连接(network type)的三种模式--bridged、host-only、NAT

    转自:http://www.cnblogs.com/xiaochaohuashengmi/archive/2011/03/15/1985084.html VMWare提供了三种工作模式,它们是brid ...

  2. [QoS]cisco3560限速配置案例-收集于网工泡泡

    网络中常用到这些:CISCO和H3C-MAC过滤+端口限速+端口镜像+端口隔离 不同的方式不同的思想:嘎嘎 其他各个厂商的限速链接:http://pan.baidu.com/s/1hrIMoSG 密码 ...

  3. CentOS 7 添加win7启动项——修改默认启动项

    CentOS 7使用grub2引导启动,在win7之后装完CentOS再启动会丢失win7启动项. 首先,添加win7启动项,步骤如下: 1.使用root登陆系统 2.用文本编辑器打开 /boot/g ...

  4. CSS 动画之十-图片+图片信息展示

    这个动画主要是运用了一些css3的特性,效果是展示一张商品图片,然后在商品图片的制定位置显示该商品的详细信息.效果在chrome浏览器中预览. <!DOCTYPE html> <ht ...

  5. Hello,cnblog‘s world!

    纠结了许久,终于选在这个地方安家.之所以选在这里,是因为这里是个很干净的社区. 之前一直在其他博文网站里发文章,可是越到后来,发现页面广告越来越多.真正对自己有价值的内容越来越少,虽然已经使用过几年了 ...

  6. [Android] 【视频】黑马安卓62、66期等教程+源码

     下载地址:http://fu83.cn/thread-58-1-1.html

  7. 20145233《Java程序设计》课程总结

    20145233 <Java程序设计>学习总结 每周学习博客汇总 20145233韩昊辰 第一周总结 20145233韩昊辰 第二周总结 20145233韩昊辰 第三周总结 2014523 ...

  8. win7 多点触摸USB的触摸屏

    USB.C 读取bCommon判断执行哪个动作 if (bCommon & rbRSUINT)//0x02 // Handle Resume interrupt { Usb_Resume(); ...

  9. Android Studio 常用快捷键及如何沿用Eclipse的快捷键

    1.显示最近编辑的文件列表 Ctrl + E 2.快速打开类 Ctrl + N 3.快速打开文件(可以是XML等各种格式后缀的文件) Ctrl + Shift + N 4.当前方法的声明 Alt + ...

  10. Java学习笔记(十八)——Java DTO

    [前面的话] 在和技术人员的交流中,各种专业术语会出现,每次都是默默的记录下出现的术语,然后再去网上查看是什么意思.最近做项目,需要使用到DTO,然后学习一下吧. 这篇文章是关于Java DTO的,选 ...