Walk Through Squares HDU

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("******
")
#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 = 13331;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1e6 + 7;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int T, m, n;
char buf[1005];
int dp[105][105][205][4];

int get_num(char ch) {
    if (ch == 'D') return 0;
    else return 1;
}

struct Aho_Corasick {
    int next[210][2], fail[210], End[210];
    int root, cnt;

    int newnode() {
        for (int i = 0; i < 2; 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][get_num(buf[i])] == -1) next[now][get_num(buf[i])] = newnode();
            now = next[now][get_num(buf[i])];
        }
        End[now] |= (1 << id);
    }

    void build() {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < 2; 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 < 2; 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]);
                }
        }
    }

    LL solve() {
        mem(dp, 0);
        dp[0][0][0][0] = 1;
        for (int i = 0; i <= n; ++i) {
            for (int j = 0; j <= m; ++j) {
                for (int k = 0; k < cnt; ++k) {
                    for (int status = 0; status < 4; ++status) {
                        if (dp[i][j][k][status] == 0) continue;
                        for (int l = 0; l < 2; ++l) {
                            int idx = next[k][l];
                            if (l == 0) {
                                dp[i + 1][j][idx][status | End[idx]] =
                                        (1LL * dp[i + 1][j][idx][status | (End[idx])] + 1LL * dp[i][j][k][status]) %
                                        mod;
//                                printf("i = %d j = %d idx = %d status = %d dp = %lld
", i + 1, j, idx,
//                                       status | (End[idx]),
//                                       dp[i + 1][j][idx][status | End[idx]]);
                            } else {
                                dp[i][j + 1][idx][status | End[idx]] =
                                        (1LL * dp[i][j + 1][idx][status | End[idx]] + 1LL * dp[i][j][k][status]) % mod;
//                                printf("i = %d j = %d idx = %d status = %d dp = %lld
", i, j + 1, idx,
//                                       status | (End[idx]),
//                                       dp[i][j + 1][idx][status | End[idx]]);
                            }
                        }
                    }
                }
            }
        }
        LL ans = 0;
        for (int i = 0; i < cnt; ++i) ans = (ans + dp[n][m][i][3]) % mod;
        return ans;
    }

    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;
    sfi(T);
    while (T--) {
        sffi(m, n);
        ac.init();
        for (int i = 0; i < 2; ++i) {
            sfs(buf);
            ac.insert(buf, i);
        }
        ac.build();
        printf("%lld
", ac.solve());
    }
    return 0;
}