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 思路是这样的：把所有病毒片段放入AC自动机中，建立fail数组。如果一个状态的fail为病毒节点，则他自己也为病毒节点。最后按边建矩阵，快速幂。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <queue>
5 using namespace std;
6 const int maxn=110;
7 const int mod=100000;
8 typedef unsigned long long ull;
9 struct Matrix{
10 int n;
11 ull mat[maxn][maxn];
12 Matrix(int n_,int on=0){
13 n=n_;memset(mat,0,sizeof(mat));
14 if(on)for(int i=1;i<=n;i++)mat[i][i]=1;
15 }
16 Matrix operator *(Matrix a){
17 Matrix ret(n);
18 unsigned long long l;
19 for(int i=1;i<=n;i++)
20 for(int k=1;k<=n;k++){
21 l=mat[i][k];
22 for(int j=1;j<=n;j++)
23 (ret.mat[i][j]+=l*a.mat[k][j]%mod)%=mod;
24 }
25 return ret;
26 }
27 Matrix operator ^(long long k){
28 Matrix ret(n,1);
29 while(k){
30 if(k&1)
31 ret=ret**this;
32 k>>=1;
33 *this=*this**this;
34 }
35 return ret;
36 }
37 };
38
39 struct AC_automation{
40 bool tag[maxn];
41 int cnt,rt,ch[maxn][4],fail[maxn];
42 AC_automation(){
43 memset(tag,0,sizeof(tag));
44 memset(fail,0,sizeof(fail));
45 memset(ch,0,sizeof(ch));cnt=rt=1;
46 }
47
48 int ID(char c){
49 if(c==‘A‘)return 0;
50 else if(c==‘C‘)return 1;
51 else if(c==‘G‘)return 2;
52 else return 3;
53 }
54
55 void Insert(char *s){
56 int len=strlen(s),p=rt;
57 for(int i=0;i<len;i++)
58 if(ch[p][ID(s[i])])
59 p=ch[p][ID(s[i])];
60 else
61 p=ch[p][ID(s[i])]=++cnt;
62 tag[p]=true;
63 }
64
65 void Build(){
66 queue<int>q;
67 for(int i=0;i<4;i++)
68 if(ch[rt][i])
69 fail[ch[rt][i]]=rt,q.push(ch[rt][i]);
70 else
71 ch[rt][i]=rt;
72
73 while(!q.empty()){
74 int x=q.front();q.pop();
75 for(int i=0;i<4;i++)
76 if(ch[x][i]){
77 fail[ch[x][i]]=ch[fail[x]][i];
78 tag[ch[x][i]]|=tag[fail[ch[x][i]]];
79 q.push(ch[x][i]);
80 }
81 else
82 ch[x][i]=ch[fail[x]][i];
83 }
84 }
85
86 void Solve(int k){
87 Matrix A(cnt);
88 for(int i=1;i<=cnt;i++)
89 for(int j=0;j<4;j++)
90 if(!tag[i]&&!tag[ch[i][j]])
91 A.mat[ch[i][j]][i]+=1;
92 A=A^k;
93 long long ans=0;
94 for(int i=1;i<=cnt;i++)
95 ans+=A.mat[i][1];
96 printf("%lld\n",ans%mod);
97 }
98 }ac;
99 char s[maxn];
100
101 int main(){
102 #ifndef ONLINE_JUDGE
103 //freopen("","r",stdin);
104 //freopen("","w",stdout);
105 #endif
106 int tot,n;
107 scanf("%d%d",&tot,&n);
108 while(tot--){
109 scanf("%s",s);
110 ac.Insert(s);
111 }
112 ac.Build();
113 ac.Solve(n);
114 return 0;
115 }