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. The Provided C++ Solution with Detailed Comments  

```cpp
#include <bits/stdc++.h>
using namespace std;

// Hungarian algorithm for finding minimum-cost perfect matching
// in a complete bipartite graph with cost matrix 'cost'.
// We will store costs in cost[0..n-1][0..n-1], and add an extra dummy
// row and column inside the class for convenience.

template<class T>
class Hungarian {
private:
    const T INF = numeric_limits<T>::max() / 2;
    int n; // number of rows (and columns)
    vector<vector<T>> cost; // 1-based inside: cost[1..n][1..n]

public:
    vector<int> assignment; // assignment[j]=i means col j matched to row i

    // Constructor: a is zero-based n×n cost matrix
    Hungarian(const vector<vector<T>>& a) {
        n = (int)a.size();
        // build an (n+1)x(n+1) matrix; we ignore row/col 0 for simplicity
        cost.assign(n+1, vector<T>(n+1, 0));
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                cost[i+1][j+1] = a[i][j];

        vector<T> u(n+1, 0), v(n+1, 0);     // potentials for rows (u) and cols (v)
        vector<int> p(n+1, 0), way(n+1, 0);// p[j]=row matched to col j; way[] for backtrace

        // For each row i, we add it to the matching
        for(int i = 1; i <= n; ++i) {
            p[0] = i;            // we will try to match row i
            int j0 = 0;          // current column we're expanding from
            vector<T> minv(n+1, INF);  // min reduced cost to each column
            vector<bool> used(n+1, false);

            // Scan columns until we find an unmatched column
            do {
                used[j0] = true;
                int i0 = p[j0];  // row currently matched to j0
                T delta = INF;
                int j1 = 0;
                // Try all columns j=1..n to improve the matching
                for(int j = 1; j <= n; ++j) {
                    if(!used[j]) {
                        // reduced cost
                        T cur = cost[i0][j] - u[i0] - v[j];
                        if(cur < minv[j]) {
                            minv[j] = cur;
                            way[j] = j0; // remember where we came from
                        }
                        if(minv[j] < delta) {
                            delta = minv[j];
                            j1 = j;
                        }
                    }
                }
                // Update potentials by delta
                for(int j = 0; j <= n; ++j) {
                    if(used[j]) {
                        u[p[j]] += delta;
                        v[j] -= delta;
                    } else {
                        minv[j] -= delta;
                    }
                }
                j0 = j1; // move to new column
            } while(p[j0] != 0); // until we reach a free column

            // Augment along the path found
            do {
                int j1 = way[j0];
                p[j0] = p[j1];
                j0 = j1;
            } while(j0 != 0);
        }

        // Build the assignment for columns 1..n
        assignment.assign(n, -1);
        for(int j = 1; j <= n; ++j) {
            assignment[j-1] = p[j] - 1; // zero-based row index
        }
    }

    // Total cost of the found matching
    T get_cost() const {
        T total = 0;
        for(int j = 0; j < n; ++j) {
            int i = assignment[j];
            total += cost[i+1][j+1];
        }
        return total;
    }
};

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

    int n;
    cin >> n;
    vector<long long> A(n);
    for(int i = 0; i < n; i++) {
        cin >> A[i];
    }

    // Read each son's preference list
    vector<vector<int>> likes(n);
    for(int i = 0; i < n; i++){
        int k; cin >> k;
        likes[i].resize(k);
        for(int j = 0; j < k; j++){
            cin >> likes[i][j];
            --likes[i][j]; // zero‐based
        }
    }

    // Build cost matrix: cost_rows = girls, cost_cols = sons
    // cost[g][s] = -A[s]^2 if son s likes girl g; else 0
    vector<vector<long long>> cost(n, vector<long long>(n, 0));
    for(int s = 0; s < n; s++){
        long long w = A[s] * A[s];
        for(int g : likes[s]){
            cost[g][s] = -w;
        }
    }

    // Solve assignment
    Hungarian<long long> hung(cost);
    auto assign = hung.assignment;  // assign[s] = g

    // Output result: if assigned girl is in son's liked list, print it+1; else 0
    for(int s = 0; s < n; s++){
        int g = assign[s];
        // check if g is actually liked
        bool ok = binary_search(likes[s].begin(), likes[s].end(), g)
                  || find(likes[s].begin(), likes[s].end(), g) != likes[s].end();
        if(ok) cout << (g+1) << ' ';
        else  cout << 0 << ' ';
    }
    cout << "\n";
    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.