AC 自动机小记
AC 自动机常用于解决多模匹配问题。
首先 \(t\) 为模式串，\(s\) 为文本串。\(Q\) 为自动机的状态集合，我们对所有模式串建立一颗字典树。记 \(fail_u\) 表示 \(v \in Q\)，\(v\) 为 \(u\) 的最长后缀。然后 \(son_{u, c}\) 表示 \(\text{trans}(u, c)\)。询问很简单，对所有点一直跳 fail 然后求和，由于所有 fail 指针组成的是一个根向树，所以可以 DFS 跑一下树上差分。时间复杂度 \(\mathcal{O}(|\Sigma|\sum |t_i| + |s_i|)\)。一些帮助理解的：ACAM 和 KMP 的匹配本质相同，每次加进来一个字符 \(c\)，如果能匹配就匹配，不能匹配就到 \(fail_u\) 上尝试，只是不断跳的过程在 build 里而已。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 2e5 + 10, M = 2e6 + 10, mod = 998244353;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
int n, son[N][26], ans[N], fail[N], idx[N], tot;
char s[M];
basic_string<int>e[N];
int insert(char s[]) {
int u = 0;
for (int i = 1; s[i]; i++) {
int c = s[i] - 'a';
if (!son[u][c]) son[u][c] = ++tot;
u = son[u][c];
}
return u;
}
void build() {
queue<int>q;
for (int i = 0; i < 26; i++) if (son[0][i]) q.push(son[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < 26; i++) {
if (son[u][i]) fail[son[u][i]] = son[fail[u]][i], q.push(son[u][i]);
else son[u][i] = son[fail[u]][i];
}
}
}
void query(char s[]) {
int u = 0;
for (int i = 1; s[i]; i++) {
u = son[u][s[i] - 'a'];
ans[u]++;
}
}
void dfs(int u) {
for (int v : e[u]) {
dfs(v);
ans[u] += ans[v];
}
}
void main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> (s + 1);
idx[i] = insert(s);
}
build();
for (int i = 1; i <= tot; i++) e[fail[i]] += i;
cin >> (s + 1);
query(s); dfs(0);
for (int i = 1; i <= n; i++) cout << ans[idx[i]] << '\n';
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
例题：
板子
- P3966 \(fail_u\) 表示的串一定会在 \(u\) 表示的串中出现，所以答案就是建出 fail 树后求子树和。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 2e5 + 10, M = 1e6 + 10, mod = 998244353;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
int n, son[M][26], ans[M], fail[M], idx[M], tot;
char s[M];
basic_string<int>e[M];
int insert(char s[]) {
int u = 0;
for (int i = 1; s[i]; i++) {
int c = s[i] - 'a';
if (!son[u][c]) son[u][c] = ++tot;
u = son[u][c];
ans[u]++;
}
return u;
}
void build() {
queue<int>q;
for (int i = 0; i < 26; i++) if (son[0][i]) q.push(son[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < 26; i++) {
if (son[u][i]) fail[son[u][i]] = son[fail[u]][i], q.push(son[u][i]);
else son[u][i] = son[fail[u]][i];
}
}
}
void query(char s[]) {
int u = 0;
for (int i = 1; s[i]; i++) {
u = son[u][s[i] - 'a'];
ans[u]++;
}
}
void dfs(int u) {
for (int v : e[u]) {
dfs(v);
ans[u] += ans[v];
}
}
void main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> (s + 1);
idx[i] = insert(s);
}
build();
for (int i = 1; i <= tot; i++) e[fail[i]] += i;
// query(s);
dfs(0);
for (int i = 1; i <= n; i++) cout << ans[idx[i]] << '\n';
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
- P2444 相当于在 ACAM 里找一个能从根走到的环，DFS 即可。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 2e5 + 10, M = 3e4 + 10, mod = 998244353;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
int n, son[M][2], b[M], vis[M], ins[M], fail[M], idx[M], tot, ok;
char s[M];
basic_string<int>e[M];
int insert(char s[]) {
int u = 0;
for (int i = 1; s[i]; i++) {
int c = s[i] - '0';
if (!son[u][c]) son[u][c] = ++tot;
u = son[u][c];
}
b[u] = 1;
return u;
}
void build() {
queue<int>q;
for (int i = 0; i < 2; i++) if (son[0][i]) q.push(son[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
b[u] |= b[fail[u]];
for (int i = 0; i < 2; i++) {
if (son[u][i]) fail[son[u][i]] = son[fail[u]][i], q.push(son[u][i]);
else son[u][i] = son[fail[u]][i];
}
}
}
void dfs(int u) {
vis[u] = ins[u] = 1;
for (int i = 0; i < 2; i++) if (son[u][i] && !b[son[u][i]]) {
if (ins[son[u][i]]) ok = 1;
else if (!vis[son[u][i]]) dfs(son[u][i]);
}
ins[u] = 0;
}
void main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> (s + 1);
idx[i] = insert(s);
}
if (!son[0][0] || !son[0][1]) {
cout << "TAK\n";
return ;
}
build();
// query(s);
dfs(0);
cout << (ok ? "TAK\n" : "NIE\n");
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
- CF1202E 首先考虑转化为枚举分界点 \(i\)，令 \(f_i\) 表示 \(s_j\) 与 \(t_{[l, i]}\) 匹配的 \(j\) 的个数，\(g_i\) 表示 \(s_j\) 与 \(t_{[i, r]}\) 匹配的 \(j\) 的个数，则答案为 \(\sum f_i \times g_{i + 1}\)，\(f\) 的话，记一下其 fail 树上到根节点的作为某字符串结尾的节点的个数。\(g\) 的话考虑翻转一下 \(s, t\) 就变成了和求 \(f\) 一样的。时间复杂度 \(\mathcal{O}(n + m)\)。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 2e5 + 10, mod = 998244353;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
int n, f[N], g[N];
string s, t;
struct ACAM {
int tot, son[N][26], fail[N], val[N];
void insert(string s) {
int u = 0;
for (char c : s) {
int &v = son[u][c - 'a'];
if (!v) v = ++tot;
u = v;
}
val[u]++;
}
void build() {
queue<int>q;
for (int i = 0; i < 26; i++) if (son[0][i]) q.push(son[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < 26; i++) {
int &v = son[u][i];
if (v) {
fail[v] = son[fail[u]][i];
val[v] += val[fail[v]];
q.push(v);
} else v = son[fail[u]][i];
}
}
}
void query(string s, int *f) {
int u = 0;
for (int i = 0; i < (int)s.size(); i++) {
u = son[u][s[i] - 'a'];
f[i] = val[u];
}
}
} acam1, acam2;
void main() {
cin >> t;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s;
acam1.insert(s);
reverse(s.begin(), s.end());
acam2.insert(s);
}
acam1.build(); acam2.build();
acam1.query(t, f);
reverse(t.begin(), t.end());
acam2.query(t, g);
ll ans = 0;
int len = (int)t.size();
for (int i = 0; i + 1 < len; i++) {
ans += (ll)f[i] * g[len - i - 2];
}
cout << ans << '\n';
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
// start coding at 19:39
// finish debugging at 20:04
ACAM 上 DP
- P4052 考虑 \(dp_{i, u}\) 表示长度为 \(i\)，ACAM 上走到了节点 \(u\)，转移显然。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 1e2 + 10, M = 6e3 + 10, mod = 1e4 + 7;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
void add(int &x, int y) { x += y; if (x >= mod) x -= mod; }
int n, m, dp[N][M], ans;
char s[N][N];
struct ACAM {
int tot, son[M][26], fail[M], b[M];
void insert(char s[]) {
int u = 0, len = strlen(s + 1);
for (int i = 1; i <= len; i++) {
int &v = son[u][s[i] - 'A'];
if (!v) v = ++tot;
u = v;
}
b[u] = 1;
}
void build() {
queue<int>q;
for (int i = 0; i < 26; i++) if (son[0][i]) q.push(son[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < 26; i++) {
int &v = son[u][i];
if (v) {
fail[v] = son[fail[u]][i];
b[v] |= b[fail[v]];
q.push(v);
} else v = son[fail[u]][i];
}
}
}
void DP() {
dp[0][0] = 1;
for (int i = 0; i < m; i++) {
for (int u = 0; u <= tot; u++) {
for (int j = 0; j < 26; j++) if (!b[son[u][j]]) {
add(dp[i + 1][son[u][j]], dp[i][u]);
}
}
}
for (int i = 0; i <= tot; i++) add(ans, dp[m][i]);
}
} acam;
void main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
cin >> (s[i] + 1);
acam.insert(s[i]);
}
acam.build();
acam.DP();
int fac = 1;
for (int i = 1; i <= m; i++) fac = 26ll * fac % mod;
cout << (fac - ans + mod) % mod << '\n';
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
// start coding at unknown
// finish debugging at 21:22
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 2e3 + 10, M = 1e2 + 10, mod = 1e9 + 7;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
int n, m, f[N][N][2][2];
char s[M][N], t[N];
struct ACAM {
int tot, son[N][10], fail[N], val[N];
void insert(char s[]) {
int u = 0, len = strlen(s + 1);
for (int i = 1; i <= len; i++) {
int &v = son[u][s[i] - '0'];
if (!v) v = ++tot;
u = v;
}
val[u] = 1;
}
void build() {
queue<int>q;
for (int i = 0; i < 10; i++) if (son[0][i]) q.push(son[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < 10; i++) {
int &v = son[u][i];
if (v) {
fail[v] = son[fail[u]][i];
val[v] |= val[fail[v]];
q.push(v);
} else v = son[fail[u]][i];
}
}
}
int dfs(int d, int u, bool lim, bool lz) {
if (d == n + 1) return !lz;
if (~f[d][u][lim][lz]) return f[d][u][lim][lz];
int res = 0, up = lim ? t[d] - '0' : 9;
for (int i = 0; i <= up; i++) {
if (lz && (i == 0)) res = (res + dfs(d + 1, 0, lim & (i == up), 1)) % mod;
else if (!val[son[u][i]]) res = (res + dfs(d + 1, son[u][i], lim & (i == up), 0)) % mod;
}
return f[d][u][lim][lz] = res;
}
} acam;
void main() {
cin >> (t + 1) >> m;
n = strlen(t + 1);
for (int i = 1; i <= m; i++) {
cin >> (s[i] + 1);
acam.insert(s[i]);
}
acam.build();
memset(f, 255, sizeof f);
cout << acam.dfs(1, 0, 1, 1) << '\n';
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
// start coding at 19:41
// finish debugging at 19:59
- P2292 考虑暴力跳 fail 的 DP：\(f_i = \lor f_j[\exists s_i = t_{[j + 1, i]}]\)，注意到 \(|s_i| \le 20\)，所以可以状压所有可以转移的位置。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 30, M = 2e6 + 10, mod = 998244353;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
int n, m;
char s[N][N], t[M];
struct ACAM {
int tot, son[N * N][26], idx[N * N], fail[N * N], dep[N * N], st[N * N];
void insert(char s[]) {
int u = 0;
for (int i = 1; s[i]; i++) {
int &v = son[u][s[i] - 'a'];
if (!v) v = ++tot;
u = v;
}
idx[u] = 1;
}
void build() {
queue<int>q;
for (int i = 0; i < 26; i++) if (son[0][i]) q.push(son[0][i]), dep[son[0][i]] = 1;
while (!q.empty()) {
int u = q.front(); q.pop();
st[u] = st[fail[u]];
if (idx[u]) st[u] |= 1 << dep[u];
for (int i = 0; i < 26; i++) {
int &v = son[u][i];
if (v) {
fail[v] = son[fail[u]][i];
dep[v] = dep[u] + 1;
q.push(v);
} else v = son[fail[u]][i];
}
}
}
int query(char t[]) {
int u = 0, mx = 0;
unsigned s = 1;
for (int i = 1; t[i]; i++) {
u = son[u][t[i] - 'a'];
s <<= 1;
if (st[u] & s) s |= 1, mx = i;
}
return mx;
}
} acam;
void main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
cin >> (s[i] + 1);
acam.insert(s[i]);
}
acam.build();
while (m--) {
cin >> (t + 1);
cout << acam.query(t) << '\n';
}
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
// start coding at 17:58
// finish debugging at 18:17
搭配 DS
- P2414 考虑子串等价于某个前缀的后缀，前缀就是 trie 上 \(u\) 到根的路径上的点，后缀就是 fail 树上的点，于是相当于 fail 树上 \(x\) 的子树与 trie 上 \(y\) 到根节点路径的交集的大小，路径加子树查，BIT 即可。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 2e5 + 10, mod = 998244353;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
char s[N];
int Q, L[N], R[N], idx, stk[N], top, a[N], cnt;
int son[N][26], fail[N], tot;
int ans[N];
vector<pii>qry[N];
basic_string<int>e[N], g[N];
struct Fenwick {
int tr[N];
void add(int x, int v) { for (int i = x; i <= idx; i += i & -i) tr[i] += v; }
int ask(int x) { int res = 0; for (int i = x; i; i -= i & -i) res += tr[i]; return res; }
int ask(int l, int r) { return ask(r) - ask(l - 1); }
} BIT;
void dfs1(int u) {
L[u] = ++idx;
for (int v : e[u]) dfs1(v);
R[u] = idx;
}
void dfs2(int u) {
BIT.add(L[u], 1);
for (auto [x, id] : qry[u]) {
ans[id] = BIT.ask(L[x], R[x]);
}
for (int v : g[u]) dfs2(v);
BIT.add(L[u], -1);
}
void main() {
cin >> (s + 1);
for (int i = 1; s[i]; i++) {
if (s[i] == 'B') top--;
else if (s[i] == 'P') a[++cnt] = stk[top];
else {
int u = stk[top], &v = son[u][s[i] - 'a'];
if (!v) v = ++tot;
stk[++top] = v;
}
}
for (int i = 0; i <= tot; i++) {
for (int j = 0; j < 26; j++) if (son[i][j]) g[i] += son[i][j];
}
queue<int>q;
for (int i = 0; i < 26; i++) if (son[0][i]) q.push(son[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
e[fail[u]] += u;
for (int i = 0; i < 26; i++) {
int &v = son[u][i];
if (v) {
fail[v] = son[fail[u]][i];
q.push(v);
} else v = son[fail[u]][i];
}
}
dfs1(0);
cin >> Q;
for (int i = 1; i <= Q; i++) {
int u, v; cin >> u >> v;
qry[a[v]].push_back({a[u], i});
}
dfs2(0);
for (int i = 1; i <= Q; i++) cout << ans[i] << '\n';
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
// start coding at 21:19
// finish debugging at 21:52
- CF163E ACAM 是个离线算法，但是这题我们可以直接初始插入，删除就是点权置 \(0\)，查询就是每个前缀 fail 树上到根节点的权值和。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 1e6 + 10, mod = 998244353;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
int n, Q, tot, idx, son[N][26], fail[N], pos[N], b[N], L[N], R[N];
string s[N], t;
basic_string<int>e[N];
struct Fenwick {
int tr[N];
void add(int x, int v) { for (int i = x; i <= idx; i += i & -i) tr[i] += v; }
void add(int l, int r, int v) { add(l, v); add(r + 1, -v); }
int ask(int x) { int res = 0; for (int i = x; i; i -= i & -i) res += tr[i]; return res; }
} BIT;
void insert(string s, int id) {
int u = 0;
for (char c : s) {
int &v = son[u][c - 'a'];
if (!v) v = ++tot;
u = v;
}
pos[id] = u;
}
void build() {
queue<int>q;
for (int i = 0; i < 26; i++) if (son[0][i]) q.push(son[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
e[fail[u]] += u;
for (int i = 0; i < 26; i++) {
int &v = son[u][i];
if (v) {
fail[v] = son[fail[u]][i];
q.push(v);
} else v = son[fail[u]][i];
}
}
}
void dfs(int u) {
L[u] = ++idx;
for (int v : e[u]) dfs(v);
R[u] = idx;
}
int query(string t) {
int u = 0;
ll res = 0;
for (char c : t) {
if (c == '?') continue;
u = son[u][c - 'a'];
res += BIT.ask(L[u]);
}
return res;
}
void main() {
cin >> Q >> n;
for (int i = 1; i <= n; i++) {
cin >> s[i];
insert(s[i], i);
}
build();
dfs(0);
for (int i = 1; i <= n; i++) {
BIT.add(L[pos[i]], R[pos[i]], 1);
b[i] = 1;
}
while (Q--) {
cin >> t;
char op = t[0];
if (op == '?') cout << query(t) << '\n';
else {
int x = 0;
for (int i = 1; i < (int)t.size(); i++) x = x * 10 + t[i] - '0';
if (op == '-' && b[x]) {
b[x] = 0;
x = pos[x];
BIT.add(L[x], R[x], -1);
}
if (op == '+' && !b[x]) {
b[x] = 1;
x = pos[x];
BIT.add(L[x], R[x], 1);
}
}
}
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
// start coding at 10:56
// finish debugging at 11:27
点击查看代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 2e6 + 10, mod = 998244353;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
cout << arg << ' ';
dbg(args...);
}
namespace Loop1st {
int n, Q;
int tot, son[N][26], fail[N], pos[N];
int idx, L[N], R[N], fa[N], top[N], hson[N], sz[N], dep[N];
string s[N], t;
vector<int>a;
basic_string<int>e[N];
struct Fenwick {
int tr[N];
void add(int x, int v) { for (int i = x; i <= idx; i += i & -i) tr[i] += v; }
int ask(int x) { int res = 0; for (int i = x; i; i -= i & -i) res += tr[i]; return res; }
int ask(int l, int r) { return ask(r) - ask(l - 1); }
} BIT;
void insert(string s, int id) {
int u = 0;
for (char c : s) {
int &v = son[u][c - 'a'];
if (!v) v = ++tot;
u = v;
}
pos[id] = u;
}
void build() {
queue<int>q;
for (int i = 0; i < 26; i++) if (son[0][i]) q.push(son[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
e[fail[u]] += u; fa[u] = fail[u];
for (int i = 0; i < 26; i++) {
int &v = son[u][i];
if (v) {
fail[v] = son[fail[u]][i];
q.push(v);
} else v = son[fail[u]][i];
}
}
}
void dfs1(int u) {
sz[u] = 1;
for (int v : e[u]) {
dep[v] = dep[u] + 1;
dfs1(v);
sz[u] += sz[v];
if (!hson[u] || sz[v] > sz[hson[u]]) hson[u] = v;
}
}
void dfs2(int u, int tp) {
L[u] = ++idx;
top[u] = tp;
if (hson[u]) dfs2(hson[u], tp);
for (int v : e[u]) if (v != hson[u]) dfs2(v, v);
R[u] = idx;
}
int lca(int u, int v) {
int x = top[u], y = top[v];
while (x != y) {
if (dep[x] < dep[y]) swap(u, v), swap(x, y);
u = fa[x];
x = top[u];
}
return dep[u] < dep[v] ? u : v;
}
void update(string t) {
a.clear();
int u = 0;
for (char c : t) {
u = son[u][c - 'a'];
BIT.add(L[u], 1);
a.push_back(u);
}
sort(a.begin(), a.end(), [&](const int &x, const int &y) { return L[x] < L[y]; });
for (int i = 0; i < (int)a.size() - 1; i++) {
BIT.add(L[lca(a[i], a[i + 1])], -1);
}
}
void main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s[i];
insert(s[i], i);
}
build();
dfs1(0);
dfs2(0, 0);
cin >> Q;
while (Q--) {
int op, x; cin >> op;
if (op == 1) {
cin >> t;
update(t);
} else {
cin >> x;
x = pos[x];
cout << BIT.ask(L[x], R[x]) << '\n';
}
}
}
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T--) Loop1st::main();
return 0;
}
// start coding at 11:50
// finish debugging at 12:04
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