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 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:

problemId=4114" id="rfc_long_toggle" style="color:blue; text-decoration:none; font-family:Arial,Helvetica,Verdana,sans-serif; font-size:15px">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 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

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.

题意：给出n个编码后的模板串，然后有M次询问，每次询问输入一个编码后的文本串。问在编码前，有多少个模板串在文本串中出现过。

分析：我觉得这道题的难点不在于构建AC自己主动机。而在于怎样把编码后的字符串解码成原文。编码时先把字符串中的每个字符转化为一个十进制整数。然后把每个整数转化为8位2进制整数。然后把这些2进制数连接起来。从头開始每次取6位，不足6位时在后边补0，然后把这6位二进制转化为一个新的十进制整数。这个新的整数就是编码后的字符的ASCII码。编码前的字符数量假设不是3的倍数，则在编码后加入‘=’，数量为字符数除以3的余数。

由于解码后的字符的ASCII值是从0~255的。所以最好不要用字符串存储解码后的内容。用int来保存较好一些。

#include<cstdio>

#include<cstring>

#include<vector>

#include<queue>

using namespace std;

const int kind = 256; //the maximum number of letter

const int N = 50010;

struct Node

{

    int next[kind];

    int fail;

    int cnt;

    void build_node() {

        for(int i = 0; i < kind; i++)

            next[i] = 0;

        fail = 0;

        cnt = 0;

    }

} node[N];

int digit[N], code[N], Size, vis[N];

char file[N], vir[N];

int fun(char ch)  //convert char to int

{

    if(ch >= 'A' && ch <= 'Z') return ch - 'A';

    if(ch >= 'a' && ch <= 'z') return ch - 'a' + 26;

    if(ch >= '0' && ch <= '9') return ch - '0' + 52;

    if(ch == '+') return 62;

    return 63;

}

void change(char *str)

{

    int i, j, len, t;

    vector<int> v;

    memset(digit, 0, sizeof(digit));

    for(len = strlen(str); str[len-1] == '='; len--) ;

    str[len] = '\0';

    for(i = 0; i < len; i++)

        v.push_back(fun(str[i]));

    for(i = 0; i < v.size(); i++) {

        for(j = 6 * (i + 1) - 1; j >= 6 * i; j--) {

            if(v[i]&1)

                digit[j] = 1;

            v[i] >>= 1;

        }

    }

    int k = v.size() * 6 / 8; //the number of letter before encode

    for(i = 0; i < k; i++) {

        for(t = 0, j = 8 * i; j < 8 * (i + 1); j++)

            t = (t << 1) + digit[j];

        code[i] = t;

    }

    code[i] = -1;

}

void Insert()

{

    int cur, index, i;

    for(cur = i = 0; code[i] >= 0; i++) {

        index = code[i];

        if(node[cur].next[index] == 0) {

            node[++Size].build_node();

            node[cur].next[index] = Size;

        }

        cur = node[cur].next[index];

    }

    node[cur].cnt++;

}

void build_ac_automation(int root)

{

    queue<int> q;

    q.push(root);

    while(!q.empty()) {

        int cur = q.front(); q.pop();

        for(int i = 0; i < kind; i++) {

            if(node[cur].next[i]) {

                int p = node[cur].next[i];

                if(cur)

                    node[p].fail = node[node[cur].fail].next[i];

                q.push(p);

            }

            else

                node[cur].next[i] = node[node[cur].fail].next[i];

        }

    }

}

int Query()

{

    int ans = 0;

    for(int i = 0, cur = 0; code[i] >= 0; i++) {

        int index = code[i];

        cur = node[cur].next[index];

        for(int j = cur; j; j = node[j].fail) {

            if(!vis[j]) {

                ans += node[j].cnt;

                vis[j] = 1;

            }

        }

    }

    return ans;

}

int main()

{

    int n, m;

    while(~scanf("%d",&n)) {

        node[0].build_node();

        Size = 0;

        for(int i = 1; i <= n; i++) {

            scanf("%s",vir);

            change(vir);

            Insert();

        }

        build_ac_automation(0);

        scanf("%d",&m);

        while(m--) {

            memset(vis, 0, sizeof(vis));

            scanf("%s", file);

            change(file);

            printf("%d\n", Query());

        }

        printf("\n");

    }

    return 0;

}