Wireless Password HDU - 2825

题意：

给出m个模式串，要求构造一长度为n的文本串，至少包括k种模式串，求有多少种可能的模式串。

k<=10  然后可以想到状压

一个文本串，k种模式串，很容易想到AC自动机。

把所有的模式串放入AC自动机上面，然后跑状压DP

跟AC自动机有关的DP一般都要用的AC自动机上的节点。

dp状态定义为dp[ i ][ j ][status]走到长度为i 时，在AC自动机上 j 这个节点

状态为 status 的方案数。

然后统计答案即可。

由于状态只与上一步有关，所以我滚动了一维，多开一维并不会MLE，也可以写。

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <time.h>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 #define  pi acos(-1.0)
 #define  eps 1e-9
 #define  fi first
 #define  se second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug               printf("******\n")
 #define  mem(a, b)         memset(a,b,sizeof(a))
 #define  name2str(x)       #x
 #define  fuck(x)           cout<<#x" = "<<x<<endl
 #define  sf(n)             scanf("%d", &n)
 #define  sff(a, b)         scanf("%d %d", &a, &b)
 #define  sfff(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  pf                printf
 #define  FIN               freopen("../date.txt","r",stdin)
 #define  gcd(a, b)         __gcd(a,b)
 #define  lowbit(x)         x&-x
 #define  IO                iOS::sync_with_stdio(false)

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const int maxn = 1e6 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = ;

 char buf[];
 int n, m, k;
 LL dp[][][];

 int cal(int num) {
     int ans = ;
     while (num) {
         if (num % ) ans++;
         num /= ;
     }
     return ans;
 }

 struct Aho_Corasick {
     int next[][], fail[], End[];
     int root, cnt;

     int newnode() {
         for (int i = ; i < ; i++) next[cnt][i] = -;
         End[cnt++] = ;
         return cnt - ;
     }

     void init() {
         cnt = ;
         root = newnode();
     }

     void insert(char buf[], int id) {
         int len = strlen(buf);
         int now = root;
         for (int i = ; i < len; i++) {
             if (next[now][buf[i] - 'a'] == -) next[now][buf[i] - 'a'] = newnode();
             now = next[now][buf[i] - 'a'];
         }
         End[now] |= ( << id);
     }

     void build() {
         queue<int> Q;
         fail[root] = root;
         for (int i = ; i < ; i++)
             if (next[root][i] == -) next[root][i] = root;
             else {
                 fail[next[root][i]] = root;
                 Q.push(next[root][i]);
             }
         while (!Q.empty()) {
             int now = Q.front();
             Q.pop();
             End[now] |=End[fail[now]];
             for (int i = ; i < ; i++)
                 if (next[now][i] == -) next[now][i] = next[fail[now]][i];
                 else {
                     fail[next[now][i]] = next[fail[now]][i];
                     Q.push(next[now][i]);
                 }
         }
     }

     void debug() {
         for (int i = ; i < cnt; i++) {
             printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
             for (int j = ; j < ; j++) printf("%2d", next[i][j]);
             printf("]\n");
         }
     }
 } ac;

 int main() {
   //   FIN;
     while (~sfff(n, m, k)) {
         if (n ==  && m ==  && k == ) break;
         ac.init();
         for (int i = ; i < m; ++i) {
             scanf("%s", buf);
             ac.insert(buf, i);
         }
         ac.build();
         for (int i = ; i <; i++)
             for (int j = ; j < ac.cnt; j++)
                 for (int p = ; p < ( << m); p++)
                     dp[i][j][p] = ;
         int now = ;
         dp[now][][] = ;
         for (int i = ; i <= n; ++i) {
             now = now ^ ;
             for (int j = ; j < ac.cnt; j++)
                 for (int p = ; p < ( << m); p++)
                     dp[now][j][p] = ;
             for (int j = ; j < ac.cnt; ++j) {
                 for (int status = ; status < ( << m); ++status) {
                     if (dp[now ^ ][j][status]) {
                         for (int l = ; l < ; ++l) {
                             int st = (status | ac.End[ac.next[j][l]]);
                             dp[now][ac.next[j][l]][st] = (dp[now][ac.next[j][l]][st] + dp[now ^ ][j][status]) % mod;
                         }
                     }
                 }
             }
         }
         LL ans = ;
         for (int status = ; status < ( << m); ++status) {
             if (cal(status) < k) continue;
             for (int i = ; i < ac.cnt; ++i)
                 ans = (ans + dp[now][i][status]) % mod;
         }
         printf("%lld\n", ans);
     }
     return ;
 }