1. Abridged Problem Statement  
Given N sons (1≤N≤400), each son i has a “love weight” A_i and a list of liked girls (girls are numbered 1…N). We wish to marry some sons to distinct girls they like (others stay unmarried) so as to maximize  
   sqrt( ∑ A_i² over married sons ).  
Output, for each son i, the girl’s index he marries (or 0 if he remains unmarried).  

2. Detailed Editorial  

Modeling as a Bipartite Matching Problem  
-----------------------------------------  
- We have a bipartite graph with N “son” vertices on the left and N “girl” vertices on the right.  
- If son i likes girl j, we may match i–j at profit A_i². Otherwise profit = 0 (or disallowed).  
- Selecting any matching yields total happiness ∑ A_i² over matched sons; taking the square root is monotonic, so we simply maximize ∑ A_i².  

Reduction to the Assignment Problem  
------------------------------------  
- The classical assignment (Hungarian) algorithm solves the problem of perfectly matching N left to N right vertices to minimize total cost on a complete bipartite graph.  
- We convert profits to costs by defining cost[i][j] = –A_j² if son j likes girl i, else cost[i][j] = 0.  
- Then a minimum-cost perfect matching on this N×N matrix will pick as many large negative costs as possible, i.e. maximize the sum of A_j².  
- Sons who cannot or should not marry any liked girl simply get matched via zero‐cost edges; we detect these afterwards and output 0.  

Hungarian Algorithm Outline  
----------------------------  
1. Build an (N+1)×(N+1) cost matrix `cost`, 0-indexed but with an extra dummy row/column for algorithmic convenience.  
2. Maintain dual potentials `u[0..n]` (for rows) and `v[0..m]` (for columns), and an array `p[0..m]` where `p[j]` is the index of the row currently matched to column j.  
3. For each row i (0…n–1), “augment” the partial matching by finding the cheapest way to assign it, updating potentials to maintain reduced costs ≥0, and then tracing back the “way” pointers to fix the matching.  
4. The resulting `p[j]` for j=0..m–1 gives the matching row for each column; invert that to get for each son (column) which girl (row) he matches to.  

Complexity  
----------  
- Hungarian runs in O(N³). With N≤400, that is ≈64·10^6 operations, which is feasible in optimized C++ (0.25 s).  

Output Reconstruction  
----------------------  
- Let `assignment[j] = i` mean that column j (son j) is assigned to row i (girl i).  
- If (i) is in son j’s liked list, output i+1; otherwise output 0.  

3. C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/graph/hungarian_algorithm.hpp>

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;
};

template<class T>
class HungarianAlgorithm {
  private:
    const T INF = numeric_limits<T>::max() / 2;
    vector<vector<T>> cost;

  public:
    vector<int> assignment;

    HungarianAlgorithm(const vector<vector<T>>& a) {
        int n = a.size(), m = a[0].size();
        cost.assign(n + 1, vector<T>(m + 1));
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                cost[i][j] = a[i][j];
            }
        }

        vector<T> u(n + 1), v(m + 1);
        vector<int> p(m + 1, n), way(m + 1, n);
        for(int i = 0; i < n; i++) {
            p[m] = i;
            int j0 = m;
            vector<T> minv(m + 1, INF);
            vector<bool> used(m + 1, false);
            do {
                used[j0] = true;
                int i0 = p[j0], j1;
                T delta = INF;
                for(int j = 0; j < m; j++) {
                    if(!used[j]) {
                        T cur = cost[i0][j] - u[i0] - v[j];
                        if(cur < minv[j]) {
                            minv[j] = cur;
                            way[j] = j0;
                        }
                        if(minv[j] < delta) {
                            delta = minv[j];
                            j1 = j;
                        }
                    }
                }
                for(int j = 0; j <= m; j++) {
                    if(used[j]) {
                        u[p[j]] += delta;
                        v[j] -= delta;
                    } else {
                        minv[j] -= delta;
                    }
                }
                j0 = j1;
            } while(p[j0] != n);

            do {
                int j1 = way[j0];
                p[j0] = p[j1];
                j0 = j1;
            } while(j0 != m);
        }

        assignment = vector<int>(begin(p), end(p) - 1);
    }

    T get_cost() {
        T ans = 0;
        for(int i = 0; i < (int)assignment.size(); i++) {
            ans += cost[assignment[i]][i];
        }
        return ans;
    }
};

int n;
vector<int> a;
vector<vector<int>> adj;

void read() {
    cin >> n;
    a.resize(n);
    cin >> a;
    adj.assign(n, {});
    for(int i = 0; i < n; i++) {
        int cnt;
        cin >> cnt;
        while(cnt--) {
            int x;
            cin >> x;
            adj[i].push_back(x - 1);
        }
    }
}

void solve() {
    // Maximizing sqrt(sum of A_i^2) over matched sons is the same as maximizing
    // the sum of A_i^2 over a matching between sons and the girls they like.
    // This is a maximum-weight matching where son i matched to a liked girl
    // contributes A_i^2. We turn it into an assignment problem on an n x n cost
    // matrix: cost[girl][son] = -A_i^2 if son i likes girl, else 0, and run the
    // Hungarian algorithm to minimize total cost (i.e. maximize matched
    // squares). After solving, son i is matched to ha.assignment[i]; if that
    // girl is not in his preference list the match is a filler 0-weight pair, so
    // we output 0 for him, otherwise the girl's number.

    vector<vector<int64_t>> cost(n, vector<int64_t>(n, 0));
    for(int i = 0; i < n; i++) {
        for(int j: adj[i]) {
            cost[j][i] = -a[i] * a[i];
        }
    }

    HungarianAlgorithm<int64_t> ha(cost);
    for(int i = 0; i < n; i++) {
        if(find(adj[i].begin(), adj[i].end(), ha.assignment[i]) ==
           adj[i].end()) {
            cout << 0 << ' ';
        } else {
            cout << ha.assignment[i] + 1 << ' ';
        }
    }
    cout << '\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. A Python Solution with Detailed Comments  

```python
import sys
def hungarian(cost):
    # cost is an n×n matrix, 0-based
    n = len(cost)
    INF = 10**30
    # We use 1-based indexing internally
    u = [0]*(n+1)
    v = [0]*(n+1)
    p = [0]*(n+1)
    way = [0]*(n+1)

    for i in range(1, n+1):
        p[0] = i
        j0 = 0
        minv = [INF]*(n+1)
        used = [False]*(n+1)
        while True:
            used[j0] = True
            i0 = p[j0]
            delta = INF
            j1 = 0
            # try to improve all columns not yet used
            for j in range(1, n+1):
                if not used[j]:
                    cur = cost[i0-1][j-1] - u[i0] - v[j]
                    if cur < minv[j]:
                        minv[j] = cur
                        way[j] = j0
                    if minv[j] < delta:
                        delta = minv[j]
                        j1 = j
            # update potentials
            for j in range(n+1):
                if used[j]:
                    u[p[j]] += delta
                    v[j]   -= delta
                else:
                    minv[j] -= delta
            j0 = j1
            # found free column?
            if p[j0] == 0:
                break

        # augmenting
        while True:
            j1 = way[j0]
            p[j0] = p[j1]
            j0 = j1
            if j0 == 0:
                break

    # build assignment: for each column j, row = p[j]
    assign = [0]*n
    for j in range(1, n+1):
        assign[j-1] = p[j]-1
    # total cost can be computed if needed
    # cost_value = sum(cost[assign[j]][j] for j in range(n))
    return assign

def main():
    data = sys.stdin.read().split()
    it = iter(data)
    n = int(next(it))
    A = [int(next(it)) for _ in range(n)]
    likes = []
    for _ in range(n):
        k = int(next(it))
        arr = [int(next(it)) - 1 for __ in range(k)]
        likes.append(arr)

    # build cost matrix: rows=girls, cols=sons
    # Windows where son s likes girl g get cost = -A[s]^2
    cost = [[0]*n for _ in range(n)]
    for s in range(n):
        w = A[s]*A[s]
        for g in likes[s]:
            cost[g][s] = -w

    assign = hungarian(cost)

    # for each son s, if assigned girl is in his list, print it+1; else 0
    # convert each likes[s] to a set for O(1) checks
    lsets = [set(lst) for lst in likes]
    out = []
    for s in range(n):
        g = assign[s]
        if g in lsets[s]:
            out.append(str(g+1))
        else:
            out.append('0')
    sys.stdout.write(" ".join(out))

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

5. Compressed Editorial  
Reduce to maximum‐weight bipartite matching with weight A_i² on edges where son i likes girl j. Use the Hungarian algorithm on an N×N cost matrix with cost[j][i] = –A_i² (and 0 elsewhere) to find a minimum-cost perfect matching in O(N³). Extract each son’s assigned girl; if it wasn’t liked, output 0.