bzoj 1559 AC自动机 + dp

思路：直接在状态图上跑dp，最后枚举一下42种一下的。。 这个枚举有点恶心。

#include<bits/stdc++.h>
#define LL long long
#define ll long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 100 + 7;
const int M = 1e7 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1000000007;

int n, m, up, top, id[N], bord[N][N];
string s[11], t[11];
bool vis[N];

bool cmp(const string &a, const string &b) {
    return a.size() > b.size();
}
bool check(const string &a, const string &b) {
    for(int i = 0; i < a.size(); i++) {
        if(a.substr(i, b.size()) == b) return true;
    }
    return false;
}

int cal(const string &a, const string &b) {
    int l1 = a.size(), l2 = b.size();
    for(int i = min(l1, l2) - 1; i >= 1; i--) {
        if(a.substr(l1 - i, i) == b.substr(0, i)) return i;
    }
    return 0;
}

struct Ac {
    int ch[N][26], val[N], f[N], tot, sz, depth[N];
    LL dp[26][101][1 << 10];
    Ac(int sz) {this->sz = sz;}
    void init() {tot = 0;};
    int newNode() {
        tot++; f[tot] = 0; val[tot] = 0;
        memset(ch[tot], 0, sizeof(ch[tot]));
        return tot;
    }
    inline int idx(char c) {return c - 'a';}
    void addStr(string &s, int id) {
        int u = 0;
        for(int i = 0; i < s.size(); i++) {
            int c = idx(s[i]);
            if(!ch[u][c]) ch[u][c] = newNode(), depth[ch[u][c]] = depth[u] + 1;
            u = ch[u][c];
        }
        val[u] |= 1 << id;
    }
    void build() {
        queue<int> que;
        for(int c = 0; c < sz; c++) {
            int v = ch[0][c];
            if(!v) ch[0][c] = 0;
            else f[v] = 0, que.push(v);
        }
        while(!que.empty()) {
            int u = que.front(); que.pop();
            val[u] |= val[f[u]];
            for(int c = 0; c < sz; c++) {
                int v = ch[u][c];
                if(!v) ch[u][c] = ch[f[u]][c];
                else f[v] = ch[f[u]][c], que.push(v);
            }
        }
    }

    void solve() {
        dp[0][0][0] = 1;
        for(int k = 0; k < n; k++) {
            for(int u = 0; u <= tot; u++) {
                for(int c = 0; c < sz; c++) {
                    int v = ch[u][c];
                    for(int s = 0; s <= up; s++) {
                         dp[k + 1][v][s | val[v]] += dp[k][u][s];
                    }
                }
            }
        }

        LL ans = 0;
        for(int u = 0; u <= tot; u++)
            ans += dp[n][u][up];
        cout << ans << '\n';

        if(ans <= 42) {
            memset(vis, true, sizeof(vis));
            sort(s, s + m, cmp);
            for(int i = 0; i < m; i++)
                for(int j = i + 1; j < m; j++)
                    if(check(s[i], s[j])) vis[j] = false;

            for(int i = 0; i < m; i++)
                if(vis[i]) id[top] = top, t[top++] = s[i];

            for(int i = 0; i < top; i++)
                for(int j = 0; j < top; j++)
                    if(i != j) bord[i][j] = cal(t[i], t[j]);

            vector<string> vec;
            do {
                string ret = t[id[0]];
                for(int i = 1; i < top; i++) {
                    int len = bord[id[i - 1]][id[i]];
                    ret += t[id[i]].substr(len, 15);
                }
                if(ret.size() == n) vec.push_back(ret);
            } while(next_permutation(id, id + top));

            sort(vec.begin(), vec.end());
            for(int i = 0; i < vec.size(); i++)
                cout << vec[i] << '\n';
        }
    }
} ac(26);

int main() {
    ac.init();
    cin >> n >> m;
    for(int i = 0; i < m; i++) {
        cin >> s[i];
        ac.addStr(s[i], i);
    }
    up = (1 << m) - 1;
    ac.build();
    ac.solve();
    return 0;
}


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