題意：給定一些串，然後讓你構造出一個長度為 m 的串，並且不包含以上串，問你有多少個。

析：很明顯，如果 m 小的話 ，直接可以用DP來解決，但是 m 太大了，我們可以認為是在AC自動機圖中，根據離散中的矩陣的冪可以表示 從 i 到 j 需要 x 步的有多少條。比如A[1][2]^5 = 10，表示從結點 1 到結點 2 走五步有10種方法，利用這種方法，就可以直接進行矩陣快速冪了。

代碼如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const int maxm = 1e5 + 10;
const int mod = 100000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
const int maxnode = 10 * 10 + 10;
const int sigma = 4;

struct Matrix{
  int n;
  int a[maxnode][maxnode];
  void clear(){ ms(a, 0); }
  friend Matrix operator * (const Matrix &lhs, const Matrix &rhs){
    Matrix res;  res.n = lhs.n;
    FOR(i, 0, lhs.n)  for(int j = 0; j < lhs.n; ++j){
      res.a[i][j] = 0;
      for(int k = 0; k < lhs.n; ++k)
        res.a[i][j] = (res.a[i][j] + (LL)lhs.a[i][k] * rhs.a[k][j]) % mod;
    }
    return res;
  }
};
Matrix x;

struct Aho{
  int ch[maxnode][sigma];
  int f[maxnode];
  bool val[maxnode];
  int sz;

  void clear(){ sz = 1;  ms(ch[0], 0); }
  inline int idx(char ch){
    if(ch == ‘A‘)  return 0;
    if(ch == ‘C‘)  return 1;
    if(ch == ‘G‘)  return 2;
    return 3;
  }

  void insert(const char *s){
    int u = 0;
    while(*s){
      int c = idx(*s);
      if(!ch[u][c]){
        ms(ch[sz], 0);
        val[sz] = 0;
        ch[u][c] = sz++;
      }
      u = ch[u][c];
      ++s;
    }
    val[u] = 1;
  }

  void getFail(){
    queue<int> q;
    f[0] = 0;
    for(int c = 0; c < sigma; ++c){
      int u = ch[0][c];
      if(u){ f[u] = 0;  q.push(u); }
    }

    while(!q.empty()){
      int r = q.front();  q.pop();
      for(int c = 0; c < sigma; ++c){
        int u = ch[r][c];
        if(!u){ ch[r][c] = ch[f[r]][c];  continue; }
        q.push(u);
        int v = f[r];
        while(v && !ch[v][c])  v = f[v];
        f[u] = ch[v][c];
        val[u] |= val[f[u]];
      }
    }
  }

  void solve(){
    for(int i = 0; i < sz; ++i)  if(!val[i]){
      for(int j = 0; j < sigma; ++j){
        int u = ch[i][j];
        if(!val[u]) ++x.a[i][u];
      }
    }
    x.n = sz;
  }
};

Aho aho;
char s[maxnode];

Matrix fast_pow(Matrix x, int n){
  Matrix res;  res.cl;
  res.n = x.n;
  for(int i = 0; i < res.n; ++i)  res.a[i][i] = 1;
  while(n){
    if(n&1)  res = res * x;
    n >>= 1;
    x = x * x;
  }
  return res;
}


int main(){
  while(scanf("%d %d", &n, &m) == 2){
    aho.cl;
    for(int i = 0; i < n; ++i){
      scanf("%s", s);
      aho.insert(s);
    }
    aho.getFail();
    x.cl;
    aho.solve();
    Matrix res = fast_pow(x, m);
    int ans = 0;
    for(int i = 0; i < res.n; ++i)
      ans = (ans + res.a[0][i]) % mod;
    printf("%d\n", ans);
  }
  return 0;
}

POJ 2778 DNA Sequence (AC自動機+DP+矩陣)