DNA Sequence（POJ2778 AC自动机dp+矩阵加速）

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DNA Sequence

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

Source

POJ Monthly--2006.03.26,dodo

裸ac自动机。。写出原dp。。构造出矩阵即可。。

  1 #include<set>
  2 #include<map>
  3 #include<queue>
  4 #include<cstdio>
  5 #include<cstdlib>
  6 #include<cstring>
  7 #include<iostream>
  8 #include<algorithm>
  9 using namespace std;
 10 const int SONS=4;
 11 const int MOD=100000;
 12 const char Hash[5]={‘ ‘,‘A‘,‘C‘,‘T‘,‘G‘};
 13 #define For(i,n) for(int i=1;i<=n;i++)
 14 #define For0(i,n) for(int i=0;i<n;i++)
 15 #define Rep(i,l,r) for(int i=l;i<=r;i++)
 16 #define Down(i,r,l) for(int i=r;i>=l;i--)
 17
 18 struct trie{
 19     int sz,ch[11*11][SONS],fn[11*11],next[11*11],q[11*11*3];
 20     trie(){sz=0;}
 21     void insert(char *s){
 22         int cur=0,len=strlen(s),id;
 23         For0(i,len){
 24             For(j,4) if(s[i]==Hash[j]){id=j-1;break;}
 25             if(!ch[cur][id]) ch[cur][id]=++sz;
 26             cur=ch[cur][id];
 27         }
 28         fn[cur]=1;
 29     }
 30     void BuildAC(){
 31         int l=0,r=0;
 32         For0(i,SONS)
 33           if(ch[0][i]) q[++r]=ch[0][i];
 34         while(l<r){
 35             int cur=q[++l];
 36             For0(i,SONS){
 37                 if(ch[cur][i]){
 38                     q[++r]=ch[cur][i];
 39                     next[ch[cur][i]]=ch[next[cur]][i];
 40                     if(fn[next[ch[cur][i]]]) fn[ch[cur][i]]=1;
 41                 }
 42                 else ch[cur][i]=ch[next[cur]][i];
 43             }
 44         }
 45     }
 46 }ac;
 47
 48 int m,n,ans,dp[121][121];
 49 char st[11];
 50
 51 struct Matrix{
 52     int A[11*11][11*11];
 53     Matrix(){memset(A,0,sizeof(A));}
 54 }Unit,Ans;
 55
 56 Matrix operator * (Matrix A,Matrix B){
 57     Matrix C;
 58     For0(i,ac.sz+1)
 59       For0(j,ac.sz+1)
 60         For0(k,ac.sz+1)
 61           C.A[i][j]=(C.A[i][j]+(long long)A.A[i][k]*B.A[k][j])%MOD;
 62     return C;
 63 }
 64
 65 void DP(){
 66     For0(i,ac.sz+1)
 67         if(!ac.fn[i])
 68           For0(k,SONS){
 69               int cur=ac.ch[i][k];
 70               if(!ac.fn[cur]) Unit.A[i][cur]++;
 71           }
 72     For0(i,ac.sz+1) Ans.A[i][i]=1;
 73     while(n){
 74         if(n&1) Ans=Ans*Unit;
 75         Unit=Unit*Unit;
 76         n>>=1;
 77     }
 78     For0(i,ac.sz+1) ans=(ans+Ans.A[0][i])%MOD;
 79     printf("%d\n",ans%MOD);
 80     /*dp[0][0]=1;
 81     For(i,n)
 82       For0(j,ac.sz+1){
 83           if(ac.fn[j]) continue;
 84           For0(k,SONS){
 85               int cur=ac.ch[j][k];
 86               if(ac.fn[cur]) continue;
 87               dp[i][cur]=(dp[i][cur]+dp[i-1][j])%MOD;
 88           }
 89       }
 90     For0(i,ac.sz+1) if(!ac.fn[i]) ans=(ans+dp[n][i])%MOD;
 91     printf("%d\n",ans);*/
 92 }
 93
 94 int main(){
 95     scanf("%d%d",&m,&n);
 96     For(i,m){
 97         scanf("%s",&st);
 98         ac.insert(st);
 99     }
100     ac.BuildAC();
101     DP();
102     return 0;
103 }