[Bzoj1030][JSOI2007]文本生成器(AC自动机&dp)

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=1030

最最最常见的多串匹配问题！题目求至少包含一个子串的方案数，则可以转化成全部方案-不包含子串的方案数。

求不包含任何字串的方案数即以所有子串建AC自动机，然后跑dp，dp[i][j]表示长度为i，在AC自动机上标号为j的点的子串方案数。

$ans=sum_{i=0}^{len-1} dp[n][i]$，len为AC自动机节点编号。

最后用总方案数-ans即为最后答案。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
typedef long long ll;
const int mod = 10007;
struct node {
    int AC[60010][26];
    int End[60010];
    int fail[60010];
    int dp[110][60010];
    int len, root;
    int newcode() {
        memset(AC[len], -1, sizeof(AC[len]));
        End[len] = 0;
        len++;
        return len - 1;
    }
    void init() {
        len = 0;
        root = newcode();
    }
    void build(string s) {
        int k = s.size();
        int now = root;
        for (int i = 0; i < k; i++) {
            if (AC[now][s[i] - 'A'] == -1)
                AC[now][s[i] - 'A'] = newcode();
            now = AC[now][s[i] - 'A'];
        }
        End[now] = 1;
    }
    void getFail() {
        queue<int>q;
        fail[root] = root;
        for (int i = 0; i < 26; i++) {
            if (AC[root][i] != -1) {
                fail[AC[root][i]] = root;
                q.push(AC[root][i]);
            }
            else
                AC[root][i] = root;
        }
        while (!q.empty()) {
            int now = q.front();
            q.pop();
            End[now] |= End[fail[now]];
            for (int i = 0; i < 26; i++) {
                if (AC[now][i] != -1) {
                    fail[AC[now][i]] = AC[fail[now]][i];
                    q.push(AC[now][i]);
                }
                else
                    AC[now][i] = AC[fail[now]][i];
            }
        }
    }
    int solve(int n) {
        for (int i = 1; i <= n; i++)
            for (int j = 0; j <= len; j++)
                dp[0][0] = 0;
        dp[0][0] = 1;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < len; j++) {
                for (int k = 0; k < 26; k++) {
                    if (End[AC[j][k]] == 1)
                        continue;
                    dp[i][AC[j][k]] += dp[i - 1][j];
                    dp[i][AC[j][k]] %= mod;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < len; i++)
            ans = (ans + dp[n][i]) % mod;
        return ans;
    }
}AC;
int main() {
    string s;
    int n, m;
    scanf("%d%d", &n, &m);
    AC.init();
    for (int i = 1; i <= n; i++) {
        cin >> s;
        AC.build(s);
    }
    AC.getFail();
    int ans = AC.solve(m);
    int sum = 1;
    for (int i = 1; i <= m; i++)
        sum = sum * 26 % mod;
    printf("%d
", (sum - ans + mod) % mod);
}