ZOJ 3430 AC自动机

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http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4114

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.

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:

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.

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 i^th line contains the number of kinds of virus detected in the i^th 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.

/**
ZOJ 3430 AC自动机
题目大意：给出一定的加密的字符串，求解码后在给定串中有多少模式串出现过。
         编码就是将原串转成二进制后取每6位进行转化成10进制后与上面表格对应的字符串，末尾不足6位补0，
         且原串个数为3k+1则在编码后的串里增加'==',若个数为3k+2则增加'=',给定的模式串也是编码后的串，所以需要进行反编码
解题思路：位运算处理，解码字符串（unsigned char没有符号位正好一个字节）然后套用AC自动机就可以了
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
int n,m;
struct Trie
{
    int next[521*64][256],end[521*64],fail[521*64];
    int root,L;
    int newnode()
    {
        for(int i=0; i<256; i++)
        {
            next[L][i]=-1;
        }
        end[L++]=-1;
        return L-1;
    }
    void init()
    {
        L=0;
        root=newnode();
    }
    void insert(unsigned char *buf,int len,int id)
    {
        int now=root;
        for(int i=0; i<len; i++)
        {
            if(next[now][buf[i]]==-1)
                next[now][buf[i]]=newnode();
            now=next[now][buf[i]];
        }
        end[now]=id;
    }
    void build()
    {
        queue<int>Q;
        fail[root]=root;
        for(int i=0; i<256; 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<256; 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]);
                }
            }
        }
    }
    bool used[520];
    int query(unsigned char buf[],int len,int n)
    {
        memset(used,false,sizeof(used));
        int now = root;
        for(int i = 0;i < len;i++)
        {
            now = next[now][buf[i]];
            int temp = now;
            while( temp!=root )
            {
                if(end[temp] != -1)
                    used[end[temp]]=true;
                temp = fail[temp];
            }
        }
        int res = 0;
        for(int i = 0;i < n;i++)
            if(used[i])
                res++;
        return res;
    }
}ac;
unsigned char s[6005],buf[2050];
char str[6005];
int tot;

unsigned char Get(char c)
{
    if(c>='A'&&c<='Z') return c-'A';
    if(c>='a'&&c<='z') return c-'a'+26;
    if(c>='0'&&c<='9') return c-'0'+52;
    if(c=='+')return 62;
    return 63;
}

void change(int len)
{
    int t=0;
    for(int i=0; i<len; i+=4)
    {
        buf[t++]=((s[i]<<2)|(s[i+1]>>4));
        if(i+2<len)
        {
            buf[t++]=((s[i+1]<<4)|(s[i+2]>>2));
        }
        if(i+3<len)
        {
            buf[t++]=((s[i+2]<<6)|s[i+3]);
        }
    }
    tot=t;
}

int main()
{
    while(~scanf("%d",&n))
    {
        ac.init();
        for(int i=0; i<n; i++)
        {
            scanf("%s",str);
            int len=strlen(str);
            while(str[len-1]=='=')len--;
            for(int j=0; j<len; j++)
            {
                s[j]=Get(str[j]);
            }
            change(len);
            ac.insert(buf,tot,i);
        }
        ac.build();
        scanf("%d",&m);
        while(m--)
        {
            scanf("%s",str);
            int len=strlen(str);
            while(str[len-1]=='=')len--;
            for(int j=0; j<len; j++)
            {
                s[j]=Get(str[j]);
            }
            change(len);
            printf("%d\n",ac.query(buf,tot,n));
        }
        printf("\n");
    }
    return 0;
}

3
YmFzZTY0
dmlydXM=
dDog
1
dGVzdDogdmlydXMu

1
QA==
2
QA==
ICAgICAgICA=

2

1
0

ZOJ 3430 AC自动机

标签：

原文地址：http://blog.csdn.net/lvshubao1314/article/details/45341895