Linear algebra (Matrix multiplication): POJ 2778 DNA Sequence

DNA Sequence

Description

It's well known that DNA Sequence are 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 could mean that the animal could have a gene Tic disease. Until now scientists has 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 are a sequence that consist of a, C, T and g,and the length of sequences are a given Integer n.

Input

First line contains (0 <= m <=), N (1 <= n <=2000000000). Here, M are 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 are not larger than 10.

Output

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

Sample Input

4 3ATACAGAA

Sample Output

36
　　
The idea is this: put all the virus fragments into the AC automaton and create a fail array. If the fail for a state is a virus node, then he is also the virus node himself. Finally, the matrix is built by edge, and the power is fast.

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <queue>5 using namespacestd;6 Const intmaxn= the;7 Const intMod=100000;8typedef unsignedLong Longull;9 structmatrix{Ten     intN; One ull MAT[MAXN][MAXN]; AMatrix (intN_,inton=0){ -N=n_;memset (Mat,0,sizeof(MAT)); -         if(ON) for(intI=1; i<=n;i++) mat[i][i]=1; the     } -Matrixoperator*(Matrix a) { - Matrix ret (n); -UnsignedLong Longl; +          for(intI=1; i<=n;i++) -              for(intk=1; k<=n;k++){ +L=Mat[i][k]; A                  for(intj=1; j<=n;j++) at(Ret.mat[i][j]+=l*a.mat[k][j]%mod)%=MoD;  -             } -         returnret;  -     } -Matrixoperator^(Long Longk) { -Matrix ret (n,1); in          while(k) { -             if(k&1) toret=ret** This; +k>>=1; -* This=* This** This;  the         } *         returnret; $     }Panax Notoginseng }; -  the structac_automation{ +     BOOLTAG[MAXN]; A     intcnt,rt,ch[maxn][4],FAIL[MAXN]; the ac_automation () { +memset (Tag,0,sizeof(tag)); -memset (fail,0,sizeof(fail)); $memset (CH,0,sizeof(CH)); cnt=rt=1; $     } -      -     intID (Charc) { the         if(c=='A')return 0; -         Else if(c=='C')return 1;Wuyi         Else if(c=='G')return 2; the         Else return 3; -     } Wu      -     voidInsert (Char*R) { About         intLen=strlen (s), p=RT; $          for(intI=0; i<len;i++) -             if(Ch[p][id (S[i])) -p=Ch[p][id (S[i]); -             Else AP=ch[p][id (S[i])]=++CNT; +tag[p]=true; the     } -      $     voidBuild () { thequeue<int>Q; the          for(intI=0;i<4; i++) the             if(Ch[rt][i]) thefail[ch[rt][i]]=Rt,q.push (Ch[rt][i]); -             Else inch[rt][i]=RT; the          the          while(!Q.empty ()) { About             intx=Q.front (); Q.pop (); the              for(intI=0;i<4; i++) the                 if(Ch[x][i]) { thefail[ch[x][i]]=Ch[fail[x]][i]; +tag[ch[x][i]]|=Tag[fail[ch[x][i]]; - Q.push (Ch[x][i]); the                 }Bayi                 Else thech[x][i]=Ch[fail[x]][i]; the         } -     } -      the     voidSolve (intk) { the Matrix A (CNT);  the          for(intI=1; i<=cnt;i++) the              for(intj=0;j<4; j + +) -                 if(!tag[i]&&!Tag[ch[i][j]]) thea.mat[ch[i][j]][i]+=1; thea=a^K; the         Long Longans=0;94          for(intI=1; i<=cnt;i++) theans+=a.mat[i][1]; theprintf"%lld\n", ans%MoD); the     }98 }ac; About CharS[MAXN]; - 101 intMain () {102 #ifndef Online_judge103     //freopen ("", "R", stdin);104     //freopen ("", "w", stdout); the #endif106     intTot,n;107scanf"%d%d",&tot,&n);108      while(tot--){109scanf"%s", s); the AC. Insert (s);111     } the AC. Build ();113 AC. Solve (n); the     return 0;  the}

Linear algebra (Matrix multiplication): POJ 2778 DNA Sequence