AC自动机（出现的模式串有多少）hdu3695

最新推荐文章于 2018-02-28 17:04:48 发布

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本文介绍了一种使用AC自动机解决计算机病毒扫描问题的方法。该方法通过为每种病毒构建模式字符串，并利用AC自动机进行高效匹配，判断程序是否被感染。文章详细解析了输入输出格式及示例，并提供了完整的AC自动机实现代码。

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Computer Virus on Planet Pandora

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 256000/128000 K (Java/Others)
Total Submission(s): 2828 Accepted Submission(s): 788

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. On
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

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 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

For each test case, print an integer K in a line meaning that the program is infected by K viruses.

Sample Input

      
       3
2
AB
DCB
DACB
3
ABC
CDE
GHI
ABCCDEFIHG
4
ABB
ACDEE
BBB
FEEE
A[2B]CD[4E]F

Sample Output

      
       0
3
2

       
        
         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’.

对所有的模式串建立AC自动机，然后去匹配就行了，TLE了好多次，要各种优化才过了，最主要的也是不明白的地方，为什么用刘汝佳的last会比fail慢，以前切实过，应该快才对啊

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=250010;
const int SIGMA_SIZE=26;
char s[5100010],str[5100010];
int N;
struct AC
{
    int ch[maxn][SIGMA_SIZE],val[maxn];
    int fail[maxn];
    int Q[maxn];
    int sz;
    void clear(){memset(ch[0],0,sizeof(ch[0]));sz=1;}
    void insert(char *s)
    {
        int u=0;
        for(int i=0;s[i];i++)
        {
            int c=s[i]-'A';
            if(!ch[u][c])
            {
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz]=0;
                ch[u][c]=sz++;
            }
            u=ch[u][c];
        }
        val[u]=1;
    }
    void getfail()
    {
        queue<int> q;
        int u=0;
        for(int i=0;i<SIGMA_SIZE;i++)
        {
            u=ch[0][i];
            if(u){fail[u]=0;q.push(u);}
        }
        while(!q.empty())
        {
            int r=q.front();q.pop();
            for(int c=0;c<SIGMA_SIZE;c++)
            {
                u=ch[r][c];
                if(!u){ch[r][c]=ch[fail[r]][c];continue;}
                q.push(u);
                int v=fail[r];
                while(v&!ch[v][c])v=fail[v];
                fail[u]=ch[v][c];
            }
        }
    }
    int find(char *s,int &n)
    {
        int u=0,ans=0;
        for(int i=0;i<n;i++)
        {
            int c=s[i]-'A';
            u=ch[u][c];
            int tmp=u;
            while(tmp&&val[tmp]!=0)
            {
                ans+=val[tmp];
                val[tmp]=0;
                tmp=fail[tmp];
            }
        }
        return ans;
    }
}ac;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        ac.clear();
        for(int i=1;i<=N;i++)
        {
            scanf("%s",s);
            ac.insert(s);
        }
        ac.getfail();
        scanf("%s",s);
        int sum=0,cnt=0,len=strlen(s);
        for(int i=0;i<len;i++)
        {
            if(s[i]>='A'&&s[i]<='Z')
            {
                if(i>0&&(s[i-1]>='0'&&s[i-1]<='9'))
                {
                    for(int j=0;j<sum;j++)str[cnt++]=s[i];
                    sum=0;
                    continue;
                }
                str[cnt++]=s[i];
            }
            if(s[i]>='0'&&s[i]<='9')sum=sum*10+s[i]-'0';
        }
        str[cnt]='\0';
        int ans=ac.find(str,cnt);
        for(int i=0,j=cnt-1;i<j;i++,j--)swap(str[i],str[j]);
        ans+=ac.find(str,cnt);
        printf("%d\n",ans);
    }
    return 0;
}