Wireless Password HDU

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("******
")
#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 + 7;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 20090717;


char buf[1010];
int n, m, k;
LL dp[2][105][1025];

int cal(int num) {
    int ans = 0;
    while (num) {
        if (num % 2) ans++;
        num /= 2;
    }
    return ans;
}

struct Aho_Corasick {
    int next[110][26], fail[110], End[110];
    int root, cnt;

    int newnode() {
        for (int i = 0; i < 26; i++) next[cnt][i] = -1;
        End[cnt++] = 0;
        return cnt - 1;
    }

    void init() {
        cnt = 0;
        root = newnode();
    }

    void insert(char buf[], int id) {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; i++) {
            if (next[now][buf[i] - 'a'] == -1) next[now][buf[i] - 'a'] = newnode();
            now = next[now][buf[i] - 'a'];
        }
        End[now] |= (1 << id);
    }

    void build() {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < 26; i++)
            if (next[root][i] == -1) 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 = 0; i < 26; i++)
                if (next[now][i] == -1) 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 = 0; i < cnt; i++) {
            printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
            for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
            printf("]
");
        }
    }
} ac;


int main() {
  //   FIN;
    while (~sfff(n, m, k)) {
        if (n == 0 && m == 0 && k == 0) break;
        ac.init();
        for (int i = 0; i < m; ++i) {
            scanf("%s", buf);
            ac.insert(buf, i);
        }
        ac.build();
        for (int i = 0; i <2; i++)
            for (int j = 0; j < ac.cnt; j++)
                for (int p = 0; p < (1 << m); p++)
                    dp[i][j][p] = 0;
        int now = 0;
        dp[now][0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            now = now ^ 1;
            for (int j = 0; j < ac.cnt; j++)
                for (int p = 0; p < (1 << m); p++)
                    dp[now][j][p] = 0;
            for (int j = 0; j < ac.cnt; ++j) {
                for (int status = 0; status < (1 << m); ++status) {
                    if (dp[now ^ 1][j][status]) {
                        for (int l = 0; l < 26; ++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 ^ 1][j][status]) % mod;
                        }
                    }
                }
            }
        }
        LL ans = 0;
        for (int status = 0; status < (1 << m); ++status) {
            if (cal(status) < k) continue;
            for (int i = 0; i < ac.cnt; ++i)
                ans = (ans + dp[now][i][status]) % mod;
        }
        printf("%lld
", ans);
    }
    return 0;
}