HDU 4758 AC自动机+状压dp

题意：

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4758
一个矩形从左上角走到右下角，向右移动用R表示，向下移动用D表示，要求满足移动序列中含有给出的两个规定子串，问一共有多少种方案？

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思路：

还是经典的AC自动机+状压DP思路，类似HDU-2825：http://blog.youkuaiyun.com/bahuia/article/details/77149003
都是一个套路。

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代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const LL MOD = 1e9 + 7;

struct ACauto {
    int next[205][2], fail[205], end[205];
    int root, sz;

    int newnode() {
        for (int i = 0; i < 2; i++)
            next[sz][i] = -1;
        end[sz++] = 0;
        return sz - 1;
    }

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

    int idx(char c) {
        if (c == 'R') return 0;
        return 1;
    }

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

} ac;

int R, D;
int dp[105][105][205][4];
const char d[] = {'R', 'D'};

void solve() {
    memset(dp, 0, sizeof(dp));
    dp[0][0][0][0] = 1;
    for (int i = 0; i <= R; i++) {
        for (int j = 0; j <= D; j++) {
            for (int k = 0; k < ac.sz; k++) {
                for (int S = 0; S < 4; S++) {
                    if (dp[i][j][k][S] <= 0) continue;
                    if (i < R) {
                        int ni = i + 1, nj = j, nk = ac.next[k][0], nS = S | ac.end[nk];
                        dp[ni][nj][nk][nS] = ((LL)dp[ni][nj][nk][nS] + dp[i][j][k][S]) % MOD;
                    }
                    //cout << i << " " << j << " " << k << " " << S << " (" << dp[i][j][k][S] << ") ---> " << ni << " " << nj << " " << nk << " " << nS <<" (" << dp[ni][nj][nk][nS] << ") " << endl;
                    if (j < D) {
                        int ni = i, nj = j + 1, nk = ac.next[k][1], nS = S | ac.end[nk];
                        dp[ni][nj][nk][nS] = ((LL)dp[ni][nj][nk][nS] + dp[i][j][k][S]) % MOD;
                    }
                    //cout << i << " " << j << " " << k << " " << S << " (" << dp[i][j][k][S] << ") ---> " << ni << " " << nj << " " << nk << " " << nS << " (" << dp[ni][nj][nk][nS] << ") " << endl;
                }
            }
        }
    }
    LL ans = 0;
    for (int i = 0; i < ac.sz; i++) {
        ans = (ans + dp[R][D][i][3]) % MOD;
    }
    printf("%I64d\n", ans);
}

char str[1005];

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &R, &D);
        ac.init();
        for (int i = 0; i < 2; i++) {
            scanf("%s", str);
            ac.insert(str, i);
        }
        ac.build();
        /*for (int i = 0; i < ac.sz; i++) {
            printf("%d : ", i);
            for (int j = 0; j < 2; j++)
                printf("%d ", ac.next[i][j]);
            printf("\n");
        }*/
        solve();
    }
    return 0;
}