【POJ2778】AC自动机+矩阵乘法

DNA Sequence

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 14758	Accepted: 5716

Description

It's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence，For example, if a animal's DNA sequence contains segment ATC then it may mean that the animal may have a genetic disease. Until now scientists have found several those segments, the problem is how many kinds of DNA sequences of a species don't contain those segments.

Suppose that DNA sequences of a species is a sequence that consist of A, C, T and G，and the length of sequences is a given integer n.

Input

First line contains two integer m (0 <= m <= 10), n (1 <= n <=2000000000). Here, m is the number of genetic disease segment, and n is the length of sequences.

Next m lines each line contain a DNA genetic disease segment, and length of these segments is not larger than 10.

Output

An integer, the number of DNA sequences, mod 100000.

Sample Input

4 3
AT
AC
AG
AA

Sample Output

36

【分析】
　　之前做过的题。  建立一个AC自动机，然后把每个字符串的最后一位所在的节点标黑。（就是把mark变为1） 然后用AC自动机建立矩阵array[i][j]表示从i走到j的方案数，然后利用矩阵乘法的性质，用快速幂求出矩阵的M次幂即可。

代码依然丑：

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<queue>
 using namespace std;
 #define Maxn 1010
 #define LL long long
 #define Mod 100000

 struct node
 {
     int x,fail;
     int son[];
     bool mark;
 }t[Maxn];int tot=;

 struct Matrix
 {
     LL a[][];
 }ay,ut;

 void upd(int x)
 {
     t[x].mark=;
     memset(t[x].son,,sizeof(t[x].son));
 }

 Matrix h;
 void clear(int id)
 {
     for(int i=;i<=tot;i++)
      for(int j=;j<=tot;j++)
       if(id==) ay.a[i][j]=;
       else h.a[i][j]=;
 }

 int m,n;
 char s[Maxn];

 void read_trie()
 {
     scanf("%s",s+);
     int len=strlen(s+);
     int now=;
     for(int i=;i<=len;i++)
     {
         int ind;
         if(s[i]=='A') ind=;
         else if(s[i]=='T') ind=;
         else if(s[i]=='C') ind=;
         else ind=;
         if(!t[now].son[ind])
         {
             t[now].son[ind]=++tot,upd(tot);
         }
         now=t[now].son[ind];
         if(i==len) t[now].mark=;
     }
 }

 queue<int > q;
 void build_AC()
 {
     while(!q.empty()) q.pop();
     q.push();
     while(!q.empty())
     {
         int now=q.front();q.pop();
         int f=t[now].fail;
         for(int i=;i<=;i++)
         {
             if(t[now].son[i])
             {
                 q.push(t[now].son[i]);
                 t[t[now].son[i]].fail=now?t[f].son[i]:;
                 if(t[t[t[now].son[i]].fail].mark) t[t[now].son[i]].mark=;
             }
             else t[now].son[i]=t[f].son[i];
             if(t[t[now].son[i]].mark!=) ay.a[now][t[now].son[i]]++;
         }
     }
 }

 Matrix mul(Matrix a,Matrix b)
 {
     clear();
     for(int i=;i<=tot;i++)
      for(int j=;j<=tot;j++)
       for(int k=;k<=tot;k++)
       {
         h.a[i][j]=(h.a[i][j]+a.a[i][k]*b.a[k][j])%Mod;

       }
     return h;
 }

 void qpow(int k)
 {
     while(k)
     {
         if(k&) ut=mul(ut,ay);
         ay=mul(ay,ay);
         k>>=;
     }
 }

 void init()
 {
     scanf("%d%d",&m,&n);
     tot=;upd();
     for(int i=;i<=m;i++)
     {
         read_trie();
     }
     clear();
     build_AC();
     for(int i=;i<=tot;i++)
      for(int j=;j<=tot;j++)
         ut.a[i][j]=i==j?:;

     qpow(n);
     LL ans=;
     for(int i=;i<=tot;i++) ans=(ans+ut.a[][i])%Mod;
     printf("%I64d\n",ans);
 }

 int main()
 {
     init();
     return ;
 }

[POJ2778]

以前的code：

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<queue>
 using namespace std;

 const int Maxn=;
 const int Mod=(int)1e5;
 char s[Maxn];
 int q[Maxn];
 long long ay[Maxn][Maxn],sum[Maxn][Maxn];
 int n,m,tot,len;

 struct node
 {
     int son[],cnt,fail,mark;
 }t[Maxn];

 void floy()
 {
     tot=;
     for(int i=;i<=Maxn;i++)
     {
         t[i].cnt=;t[i].mark=;
         for(int j=;j<=;j++) t[i].son[j]=;
     }
     memset(q,,sizeof(q));
 }

 void read_trie()
 {
     tot=;
     int i,j;
     scanf("%d%d",&m,&n);
     for(i=;i<=m;i++)
     {
         int x,ind;
         scanf("%s",s+);
         len=strlen(s+);
         x=;
         for(j=;j<=len;j++)
         {
             if(s[j]=='A') ind=;
             else if(s[j]=='C') ind=;
             else if(s[j]=='T') ind=;
             else if(s[j]=='G') ind=;
             if(!t[x].son[ind]) t[x].son[ind]=++tot;
             x=t[x].son[ind];
             t[x].cnt++;
             if(j==len) t[x].mark=;
         }
     }
 }

 void build_AC()
 {
     int i,j,x,y;
     q[++q[]]=;
     //for(i=1;i<=4;i++)
       //if(t[0].son[i]) q[++q[0]]=t[0].son[i];
     for(i=;i<=q[];i++)
     {
         x=q[i];
         y=t[x].fail;
         for(j=;j<=;j++)
         {
             if(t[x].son[j])
             {
                 //t[t[x].son[j]].fail=t[y].son[j];
                 q[++q[]]=t[x].son[j];
                  t[t[x].son[j]].fail=x?t[y].son[j]:x;
                 if(t[t[t[x].son[j]].fail].mark) t[t[x].son[j]].mark=;
             }
             else t[x].son[j]=t[y].son[j];
             if(!t[t[x].son[j]].mark) ay[x][t[x].son[j]]++;
         }
     }
 }

 void MatrixMult(long long a[Maxn][Maxn],long long b[Maxn][Maxn])
 {
     int i,j,k;
     long long c[Maxn][Maxn];
     memset(c,,sizeof(c));
     for(i=;i<=tot;i++)
      for(j=;j<=tot;j++)
       for(k=;k<=tot;k++)
        c[i][j]+=a[i][k]*b[k][j];
     for(i=;i<=tot;i++)
      for(j=;j<=tot;j++)
       a[i][j]=c[i][j]%Mod;
 }

 long long MatrixPow(int k)
 {
     int i,j;
     for(i=;i<=tot;i++)
      for(j=;j<=tot;j++)
        sum[i][j]=(i==j);
     while(k)
     {
         if(k&) MatrixMult(sum,ay);
         MatrixMult(ay,ay);
         k>>=;
     }
     long long ans=;
     for(i=;i<=tot;i++) ans+=sum[][i];
     return ans%Mod;
 }

 int main()
 {
     floy();
     read_trie();
     build_AC();
     printf("%I64d\n",MatrixPow(n));
     return ;
 }

[POJ2778_2]

2016-06-23 16:22:32