Walk Through Squares HDU - 4758 AC自动机+简单状压DP

题意：给你两个串，求用m个R，n个D能组成多少个包含这两个串

题解：先构造一个AC自动机记录每个状态包含两个串的状态，

状态很容易定义 dp【i】【j】【k】【status】表示在AC自动机K这个节点

使用了 i 个D,j个R ，状态为status的方案数。

然后直接DP即可。

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <unordered_map>

 #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  sfi(a)             scanf("%d", &a)
 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  sfL(a)             scanf("%lld", &a)
 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 #define  sfs(a)             scanf("%s", a)
 #define  sffs(a, b)         scanf("%s %s", a, b)
 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 #define  FIN                freopen("../in.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 ULL seed = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxn = 1e6 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;
 int T, m, n;
 char buf[];
 int dp[][][][];

 int get_num(char ch) {
     if (ch == 'D') return ;
     else return ;
 }

 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][get_num(buf[i])] == -) next[now][get_num(buf[i])] = newnode();
             now = next[now][get_num(buf[i])];
         }
         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]);
                 }
         }
     }

     LL solve() {
         mem(dp, );
         dp[][][][] = ;
         for (int i = ; i <= n; ++i) {
             for (int j = ; j <= m; ++j) {
                 for (int k = ; k < cnt; ++k) {
                     for (int status = ; status < ; ++status) {
                         if (dp[i][j][k][status] == ) continue;
                         for (int l = ; l < ; ++l) {
                             int idx = next[k][l];
                             if (l == ) {
                                 dp[i + ][j][idx][status | End[idx]] =
                                         (1LL * dp[i + ][j][idx][status | (End[idx])] + 1LL * dp[i][j][k][status]) %
                                         mod;
 //                                printf("i = %d j = %d idx = %d status = %d dp = %lld\n", i + 1, j, idx,
 //                                       status | (End[idx]),
 //                                       dp[i + 1][j][idx][status | End[idx]]);
                             } else {
                                 dp[i][j + ][idx][status | End[idx]] =
                                         (1LL * dp[i][j + ][idx][status | End[idx]] + 1LL * dp[i][j][k][status]) % mod;
 //                                printf("i = %d j = %d idx = %d status = %d dp = %lld\n", i, j + 1, idx,
 //                                       status | (End[idx]),
 //                                       dp[i][j + 1][idx][status | End[idx]]);
                             }
                         }
                     }
                 }
             }
         }
         LL ans = ;
         for (int i = ; i < cnt; ++i) ans = (ans + dp[n][m][i][]) % mod;
         return ans;
     }

     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;
     sfi(T);
     while (T--) {
         sffi(m, n);
         ac.init();
         for (int i = ; i < ; ++i) {
             sfs(buf);
             ac.insert(buf, i);
         }
         ac.build();
         printf("%lld\n", ac.solve());
     }
     return ;
 }