DNA Sequence
| Time Limit: 1000MS | Memory Limit: 65536K | |
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.
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.
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
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 ",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 ",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 }