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。
1 #include <set> 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include <string> 8 #include 9 #include 10 #include 11 #include 12 #include 13 14 #define pi acos(-1.0) 15 #define eps 1e-9 16 #define fi first 17 #define se second 18 #define rtl rt<<1 19 #define rtr rt<<1|1 20 #define bug printf("******\n") 21 #define mem(a, b) memset(a,b,sizeof(a)) 22 #define name2str(x) #x 23 #define fuck(x) cout<<#x" = "< 24 #define sf(n) scanf("%d", &n) 25 #define sff(a, b) scanf("%d %d", &a, &b) 26 #define sfff(a, b, c) scanf("%d %d %d", &a, &b, &c) 27 #define sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d) 28 #define pf printf 29 #define FIN freopen("../date.txt","r",stdin) 30 #define gcd(a, b) __gcd(a,b) 31 #define lowbit(x) x&-x 32 #define IO iOS::sync_with_stdio(false) 33 34 35 using namespace std; 36 typedef long long LL; 37 typedef unsigned long long ULL; 38 const int maxn = 1e6 + 7; 39 const int maxm = 8e6 + 10; 40 const int INF = 0x3f3f3f3f; 41 const int mod = 20090717; 42 43 44 char buf[1010]; 45 int n, m, k; 46 LL dp[2][105][1025]; 47 48 int cal(int num) { 49 int ans = 0; 50 while (num) { 51 if (num % 2) ans++; 52 num /= 2; 53 } 54 return ans; 55 } 56 57 struct Aho_Corasick { 58 int next[110][26], fail[110], End[110]; 59 int root, cnt; 60 61 int newnode() { 62 for (int i = 0; i < 26; i++) next[cnt][i] = -1; 63 End[cnt++] = 0; 64 return cnt - 1; 65 } 66 67 void init() { 68 cnt = 0; 69 root = newnode(); 70 } 71 72 void insert(char buf[], int id) { 73 int len = strlen(buf); 74 int now = root; 75 for (int i = 0; i < len; i++) { 76 if (next[now][buf[i] - 'a'] == -1) next[now][buf[i] - 'a'] = newnode(); 77 now = next[now][buf[i] - 'a']; 78 } 79 End[now] |= (1 << id); 80 } 81 82 void build() { 83 queue<int> Q; 84 fail[root] = root; 85 for (int i = 0; i < 26; i++) 86 if (next[root][i] == -1) next[root][i] = root; 87 else { 88 fail[next[root][i]] = root; 89 Q.push(next[root][i]); 90 } 91 while (!Q.empty()) { 92 int now = Q.front(); 93 Q.pop(); 94 End[now] |=End[fail[now]]; 95 for (int i = 0; i < 26; i++) 96 if (next[now][i] == -1) next[now][i] = next[fail[now]][i]; 97 else { 98 fail[next[now][i]] = next[fail[now]][i]; 99 Q.push(next[now][i]); 100 } 101 } 102 } 103 104 void debug() { 105 for (int i = 0; i < cnt; i++) { 106 printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]); 107 for (int j = 0; j < 26; j++) printf("%2d", next[i][j]); 108 printf("]\n"); 109 } 110 } 111 } ac; 112 113 114 int main() { 115 // FIN; 116 while (~sfff(n, m, k)) { 117 if (n == 0 && m == 0 && k == 0) break; 118 ac.init(); 119 for (int i = 0; i < m; ++i) { 120 scanf("%s", buf); 121 ac.insert(buf, i); 122 } 123 ac.build(); 124 for (int i = 0; i <2; i++) 125 for (int j = 0; j < ac.cnt; j++) 126 for (int p = 0; p < (1 << m); p++) 127 dp[i][j][p] = 0; 128 int now = 0; 129 dp[now][0][0] = 1; 130 for (int i = 1; i <= n; ++i) { 131 now = now ^ 1; 132 for (int j = 0; j < ac.cnt; j++) 133 for (int p = 0; p < (1 << m); p++) 134 dp[now][j][p] = 0; 135 for (int j = 0; j < ac.cnt; ++j) { 136 for (int status = 0; status < (1 << m); ++status) { 137 if (dp[now ^ 1][j][status]) { 138 for (int l = 0; l < 26; ++l) { 139 int st = (status | ac.End[ac.next[j][l]]); 140 dp[now][ac.next[j][l]][st] = (dp[now][ac.next[j][l]][st] + dp[now ^ 1][j][status]) % mod; 141 } 142 } 143 } 144 } 145 } 146 LL ans = 0; 147 for (int status = 0; status < (1 << m); ++status) { 148 if (cal(status) < k) continue; 149 for (int i = 0; i < ac.cnt; ++i) 150 ans = (ans + dp[now][i][status]) % mod; 151 } 152 printf("%lld\n", ans); 153 } 154 return 0; 155 }
