## 1) Concise, abridged problem statement

There are **N resources** (1..N) and **M users** (1..M).  
Each user *i* requests access to a set of resources \(S_i\).

We must create some **user groups**. Each group contains:
- a set of users \(U_g\)
- a set of resources \(R_g\)

Rules:
1. **A resource can belong to at most one group** (groups’ resource sets are disjoint).
2. A user can access exactly the resources that are in groups they belong to.
3. Every user must get access to **all** requested resources and to **no other** resources.
4. A user may belong to multiple groups.

Compute the **minimum number of groups** needed.

Constraints: \(1 \le N, M \le 100\).

---

## 2) Detailed editorial (explanation of the solution)

### Key observation
Consider a particular resource \(r\). Let:

\[
U(r) = \{\, u \mid \text{user } u \text{ requests resource } r \,\}
\]

So \(U(r)\) is the set of users who must have access to resource \(r\).

Because of rule (3) (“no extra access”), **any user who does not request \(r\) must not gain access to \(r\)**. The only way to access \(r\) is to be in the group that contains \(r\) (rule 2). Therefore:

- If resource \(r\) is placed into some group \(g\), then the users in that group **must be exactly** the users that need \(r\):
  \[
  U_g = U(r)
  \]
  If \(U_g\) were bigger, extra users would gain access to \(r\).  
  If it were smaller, some required users would miss access.

So **each resource dictates the exact user set of its group**.

### When can two resources share the same group?
Rule (1) allows a resource to belong to only one group, but a group can contain multiple resources. If we put two resources \(r_1\) and \(r_2\) into the same group \(g\), then:

- Users in \(g\) get access to **all resources in that group**, including both \(r_1\) and \(r_2\).
- Therefore, the users who need \(r_1\) must be the same as the users who need \(r_2\), otherwise someone gets extra access or lacks access.

Thus:

\[
r_1 \text{ and } r_2 \text{ can be in the same group } \iff U(r_1) = U(r_2)
\]

### Minimizing the number of groups
To minimize groups, we should:
- Put **all resources with the same user-set \(U(r)\)** into one group.
- Resources with different \(U(r)\) **must** be in different groups.

Therefore, the minimal number of groups equals:

> The number of **distinct** non-empty sets \(U(r)\) over all resources \(r\).

(If a resource is requested by nobody, \(U(r)=\emptyset\). Such a resource doesn’t need to be in any group at all, so we ignore empty sets.)

### Algorithm
1. Build for each resource \(r\) the list of users requesting it.
2. For each resource with a non-empty list:
   - sort the user list (to get a canonical representation)
   - insert it into a set of vectors (deduplicates identical lists)
3. Answer = size of that set.

### Complexity
- \(N, M \le 100\). Total input size is small.
- Building lists: \(O(\text{total requests}) \le 10^4\).
- Sorting each resource’s user list: at most \(N\) lists, each length ≤ \(M\): \(O(N \cdot M \log M)\), tiny for 100.
- Using a `set<vector<int>>` is fine at this scale.

---

## 3) C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, m;
vector<vector<int>> resource_users;

void read() {
    cin >> n >> m;
    resource_users.resize(n + 1);

    for(int u = 1; u <= m; u++) {
        int k;
        cin >> k;
        for(int j = 0; j < k; j++) {
            int r;
            cin >> r;
            resource_users[r].push_back(u);
        }
    }
}

void solve() {
    // The solution is fairly simple - we can just check for every file the set
    // of people that should have access to it. We don't want any other person
    // to have access to the same file, so this corresponds to a group. The only
    // way to share groups is if there is another file with the same set of
    // people. We can simply sort the ids of the people that should have access
    // to every file, and then sort all of these - the number of groups will be
    // the number of unique vectors.

    set<vector<int>> unique_groups;
    for(int r = 1; r <= n; r++) {
        if(!resource_users[r].empty()) {
            sort(resource_users[r].begin(), resource_users[r].end());
            unique_groups.insert(resource_users[r]);
        }
    }

    cout << unique_groups.size() << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```python
import sys

def main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)

    # Read number of resources (N) and users (M)
    n = int(next(it))
    m = int(next(it))

    # resource_users[r] will store a list of users who request resource r
    # Use 1-based indexing: resources are 1..n
    resource_users = [[] for _ in range(n + 1)]

    # For each user u, read their requested resources
    for u in range(1, m + 1):
        k = int(next(it))                 # number of requested resources
        for _ in range(k):
            r = int(next(it))             # resource id
            resource_users[r].append(u)   # user u requests resource r

    # We need the number of distinct, non-empty user-sets U(r).
    # Convert each non-empty list to a sorted tuple and put into a set.
    unique_groups = set()
    for r in range(1, n + 1):
        if resource_users[r]:                     # ignore empty: nobody requests it
            users = tuple(sorted(resource_users[r]))  # canonical representation
            unique_groups.add(users)              # deduplicate equal user sets

    # The answer is the number of unique user sets
    print(len(unique_groups))

if __name__ == "__main__":
    main()
```

---

## 5) Compressed editorial

For each resource \(r\), define \(U(r)\) = users who request \(r\).  
If resource \(r\) is in some group, the group’s user set must be exactly \(U(r)\); otherwise some user gains/loses access, violating constraints. Two resources can be in the same group iff their \(U(r)\) sets are identical. Therefore the minimum number of groups equals the number of **distinct non-empty** sets \(U(r)\). Compute all \(U(r)\), sort each list, insert into a set, and output the set size.