Aho-Corasick Automaton

Exercises

Luogu P5357 - AC Automaton

CF1400F - x-prime Substrings

Hint 1

We can enumerate all x-prime strings via DFS.

Hint 2

We can build an Aho-Corasick automaton with the generated strings.

Hint 3

Consider two types of transitions:

• Discard the current digit, and increase the cost by $1$.
• Keep the current digit, and move on the Aho-Corasick automaton. This transition is prohibited if the target position is marked.

In this specific problem, no $x$-prime string is a suffix of another $x$-prime string, otherwise they will have different digit sums. In consequence, we do not need to follow fail pointers when moving on the automaton. However, do note that this is seldom the case, or you might get WA on other ACA problems (like me on ADAJOBS).

Code (C++)

#include <cstring>
#include <iostream>
#include <queue>
#include <set>
#include <vector>

using namespace std;
const int INF = 0x3f3f3f3f;

struct Node {
  int fail = 0, children[9]{};
  bool match = false;
};

string s, t;
int x, dp[5005][5005];
set<string> xprime;
vector<int> suffix = {0};
vector<Node> nodes;

void generate_xprime() {
  if (suffix[0] == x) {
    xprime.insert(t);
    return;
  }
  for (int i = 1; i <= 9 && i + suffix[0] <= x; ++i) {
    t.push_back(i + '0');
    for (int &j : suffix)
      j += i;
    suffix.emplace_back(i);
    bool ok = true;
    for (int &j : suffix)
      if (j != x && x % j == 0) {
        ok = false;
        break;
      }
    if (ok)
      generate_xprime();
    suffix.pop_back();
    for (int &j : suffix)
      j -= i;
    t.pop_back();
  }
}

void build_aca() {
  nodes.emplace_back(Node{});
  for (const string &str : xprime) {
    int curr = 0;
    for (const char &c : str) {
      if (!nodes[curr].children[c - '1']) {
        nodes[curr].children[c - '1'] = nodes.size();
        nodes.emplace_back(Node{});
      }
      curr = nodes[curr].children[c - '1'];
    }
    nodes[curr].match = true;
  }
  queue<int> q;
  for (const int &u : nodes[0].children)
    if (u)
      q.push(u);

  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < 9; ++i) {
      int &v = nodes[u].children[i];
      if (v) {
        nodes[v].fail = nodes[nodes[u].fail].children[i];
        q.push(v);
      } else
        v = nodes[nodes[u].fail].children[i];
    }
  }
}

int main() {
  cin >> s >> x;

  // Step 1: Enumerate all x-prime strings
  generate_xprime();

  // Step 2: Build Aho-Corasick automaton with x-prime strings
  build_aca();

  // Step 3: Dynamic programming
  memset(dp, 0x3f, sizeof(dp));
  int n = s.size(), m = nodes.size(), ans = INF;
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      if (dp[i][j] == INF)
        continue;
      dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + 1);
      int nxt = nodes[j].children[s[i] - '1'];
      if (!nodes[nxt].match)
        dp[i + 1][nxt] = min(dp[i + 1][nxt], dp[i][j]);
    }
  for (int j = 0; j < m; ++j)
    ans = min(ans, dp[n][j]);
  cout << ans;
}

SPOJ - Ada and Jobs

The challenge lies in that tasks are added dynamically, but we cannot afford reconstructing the whole automaton every time a new word is added.

Hint 1

Store all operations offline. When constructing the automaton, assign every stop node a timestamp indicating its order. Meanwhile, store the number of words in the dictionary at the time a query is made. In this way, we only need to build the ACA once. During matching, we need to find a stop node whose timestamp is no larger than the timestamp of the query.

Hint 2

During matching, we do not need to go all the way up along the fail pointers. Instead, we can push down the timestamp during construction.

Code (C++)

#include <iostream>
#include <queue>
#include <vector>
#define INF 0x3f3f3f3f

using namespace std;

struct Node {
  int occur = INF, fail = 0, children[26]{};
};

int main() {
  int q;
  cin >> q;
  vector<string> dict;
  vector<pair<int, string>> query;
  while (q--) {
    int t;
    string s;
    cin >> t >> s;
    if (t == 0)
      dict.emplace_back(s);
    else
      query.emplace_back(dict.size(), s);
  }
  int m = dict.size();
  vector<Node> nodes(1);
  for (int i = 1; i <= m; ++i) {
    int current = 0;
    for (char c : dict[i - 1]) {
      if (!nodes[current].children[c - 'a']) {
        nodes[current].children[c - 'a'] = nodes.size();
        nodes.emplace_back(Node{});
      }
      current = nodes[current].children[c - 'a'];
    }
    nodes[current].occur = min(nodes[current].occur, i);
  }
  queue<int> que;
  for (const int &u : nodes[0].children)
    if (u)
      que.push(u);
  while (!que.empty()) {
    int u = que.front();

    // Push down the timestamp. This is critical.
    nodes[u].occur = min(nodes[u].occur, nodes[nodes[u].fail].occur);
    que.pop();
    for (int i = 0; i < 26; ++i) {
      int &v = nodes[u].children[i];
      if (v) {
        nodes[v].fail = nodes[nodes[u].fail].children[i];
        que.push(v);
      } else
        v = nodes[nodes[u].fail].children[i];
    }
  }
  string ans;
  for (auto &[ts, s] : query) {
    int current = 0;
    bool found = false;
    int idx = 0;
    while (idx < s.size()) {
      char c = s[idx];
      if (nodes[current].children[c - 'a']) {
        current = nodes[current].children[c - 'a'];
        if (nodes[current].occur <= ts) {
          ans += "YES\n";
          found = true;
          break;
        }
        idx++;
      } else {
        current = nodes[current].fail;
        if (!current)
          idx++;
      }
    }
    if (!found)
      ans += "NO\n";
  }
  cout << ans;
}