"Multidimensional spaces are completely out of style these days, unlike genetics problems" — thought physicist Woll and changed his subject of study to bioinformatics. Analysing results of sequencing he faced the following problem concerning DNA sequences. We will further think of a DNA sequence as an arbitrary string of uppercase letters "A", "C", "G" and "T" (of course, this is a simplified interpretation).
Let w be a long DNA sequence and s1, s2, ..., sm — collection of short DNA sequences. Let us say that the collection filters w iff w can be covered with the sequences from the collection. Certainly, substrings corresponding to the different positions of the string may intersect or even cover each other. More formally: denote by |w| the length of w, let symbols of w be numbered from 1 to |w|. Then for each position i in w there exist pair of indices l, r (1 ≤ l ≤ i ≤ r ≤ |w|) such that the substring w[l ... r] equals one of the elements s1, s2, ..., sm of the collection.
Woll wants to calculate the number of DNA sequences of a given length filtered by a given collection, but he doesn't know how to deal with it. Help him! Your task is to find the number of different DNA sequences of length n filtered by the collection {si}.
Answer may appear very large, so output it modulo 1000000009.
First line contains two integer numbers n and m (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10) — the length of the string and the number of sequences in the collection correspondently.
Next m lines contain the collection sequences si, one per line. Each si is a nonempty string of length not greater than 10. All the strings consist of uppercase letters "A", "C", "G", "T". The collection may contain identical strings.
Output should contain a single integer — the number of strings filtered by the collection modulo 1000000009 (109 + 9).
2 1
A
1
6 2
CAT
TACT
2
In the first sample, a string has to be filtered by "A". Clearly, there is only one such string: "AA".
In the second sample, there exist exactly two different strings satisfying the condition (see the pictures below).
题意：简单说一下吧，就是让你构造一个长度为n的字符串，在总共m个子串中选择，构造的字符串中的每一个字符都需要有子串提供，求构成的方案。
题解：我写这题也是煞费苦心啊，写了两个晚上，重构了n次，ac自动机掌握的不扎实啊。
讲一下我的思考过程吧，也就是我抄题解的过程。
首先，由于串的匹配，并且有多个串，就很容易（？？）想到ac自动机这个算法。
然后既然是dp专题里的，就是用dp去做（formally这些数方案的差不多都是dp。。）
我们用区间dp的套路设 dp[i][j]表示匹配到i这个位置，选j的方案。发现并不能成功，于是我们再把j改成ac自动机上的节点，
于是方程就变成了dp[i][j]表示匹配到串i，到达ac自动机节点j的方案数。
然后会发现这样并不是最优解，因为串和串之间是有可能重叠的。
我们再设dp[i][j][k]表示。。。后面有k个剩余的答案。剩余表示这个串的后k个还没有匹配，接下来有可能有串匹配
于是我们非常容易推导出dp方程
dp[i][j][k]->dp[i+1][j][0]剩余k->0匹配上了。
dp[i][j][k]->dp[i+1][j][k+1]没有匹配上。
于是答案就是sigma dp[n][所有ac自动机的节点][0]
边界dp[0][0][0]=1;
于是我们就在o(100nm)的时间内解决了这个问题。
放上臭但是不长的代码。
#include<bits/stdc++.h>
#define int long long
#define N 1100
using namespace std;
const int mod =1e9+9;
int f[N][N][11],n,m,ch[N][4],val[N],fail[N],sz,root=0,mx=0;
void add(int &x,int y)
{
x+=y;
if(x<0) x+=mod;
while(x>=mod) x-=mod;
}
int idx(char c)
{
if(c=='A') return 0;
if(c=='C') return 1;
return c=='T'?3:2;
}
int newnode()
{
for(int i=0;i<4;i++) ch[sz][i]=0;
val[sz++]=0;return sz-1;
}
void insert(char *s,int len)
{
int u=0;
for(int i=0;i<len;i++)
{
int v=idx(s[i]);
if(ch[u][v]==0) ch[u][v]=newnode();
u=ch[u][v];
}
val[u]=max(val[u],len);
}
void build()
{
queue<int> q;
for(int i=0;i<4;i++)
{
if(ch[root][i]==0) ch[root][i]=root;
else fail[ch[root][i]]=root,q.push(ch[root][i]);
}
while(!q.empty())
{
int u=q.front();q.pop();
for(int i=0;i<4;i++)
{
int &v=ch[u][i];
if(v==0) v=ch[fail[u]][i];
else fail[v]=ch[fail[u]][i],q.push(v),val[v]=max(val[v],val[fail[v]]);
}
}
}
void dp()
{
f[0][0][0]=1;
for(int i=0;i<n;i++)
for(int j=0;j<=sz;j++)
for(int k=0,v;k<=mx;k++)if(v=f[i][j][k])
{
//printf ( "dp[%d][%d][%d] = %d\n" , i , j , k , f[i][j][k] );
for(int p=0;p<4;p++)
{
int now=ch[j][p];
if(val[now]>k) add(f[i+1][now][0],v);
else if(k<mx) add(f[i+1][now][k+1],v);
}
}
int ans=0;
for(int i=0;i<=sz;i++) add(ans,f[n][i][0]);
cout<<ans<<"\n";
// for(int i=1;i<=sz;i++) cout<<val[i]<<"\n";
}
char str[200];
main()
{
//freopen("1.txt","w",stdout);
root=newnode();
cin>>n>>m;
for(int i=1,len;i<=m;i++)
scanf("%s",str),len=strlen(str),mx=max(mx,len),insert(str,len);
build();
dp();
return 0;
}