1. Abridged Problem Statement

We have an undirected graph of N cities and M roads. Roads 1 through N–1 form a spanning tree (the “stoned” roads). Each road i has a true maintenance cost cᵢ. We must report fake costs dᵢ so that, in the graph with weights dᵢ, the unique minimum spanning tree (ties allowed) is exactly the original tree of roads 1…N–1. Among all such choices of d, we want to minimize ∑|cᵢ–dᵢ|.

2. Detailed Editorial

We want to adjust each tree edge’s cost upward (by xᵢ ≥ 0) and each non-tree edge’s cost downward (by yⱼ ≥ 0), so that no non-tree edge can replace a tree edge in the MST. Concretely, if a non-tree edge j (with true cost cⱼ) connects vertices u and v, its path in the tree goes through some tree edges i. We require for every such pair (i, j):

  cᵢ + xᵢ  ≤  cⱼ – yⱼ.

Equivalently,

  xᵢ + yⱼ  ≤  cⱼ – cᵢ.

Our goal is to choose nonnegative xᵢ, yⱼ to satisfy these constraints and to minimize

  ∑_{i in tree} xᵢ  +  ∑_{j not in tree} yⱼ.

This is a classic linear program with a totally unimodular constraint matrix, whose dual is a maximum‐weight feasible potential problem on a complete bipartite graph between “tree edges” and “all edges.” We can solve it by the Hungarian algorithm (Kuhn–Munkres) on an (n–1)×M cost matrix A where

  A[i][j] = (cⱼ – cᵢ)   if tree edge i lies on the path of edge j,
            0           otherwise.

The Hungarian algorithm computes dual potentials u[i] (for tree edges) and v[j] (for all edges) that maximize ∑u[i] + ∑v[j] subject to

  u[i] + v[j]  ≤  A[i][j]   for all i,j.

It turns out that the optimal u[i], v[j] exactly equal the chosen increments xᵢ = u[i] and decrements yⱼ = v[j]. Finally we report

  dᵢ = cᵢ + xᵢ   for i in tree (1…N–1),
  dⱼ = cⱼ – yⱼ   for j not in tree (N…M).

Overall complexity is O(N·M·(N + M)) which is fine for N≤60, M≤400.

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<int> way;

  public:
    int n, m;
    vector<vector<T>> cost;
    vector<int> assignment;
    vector<T> pot_left, pot_right;

    HungarianAlgorithm(const vector<vector<T>>& a) {
        n = a.size();
        m = a[0].size();
        assert(n <= m);

        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];
            }
        }

        pot_left.assign(n + 1, 0);
        pot_right.assign(m + 1, 0);
        assignment.assign(m + 1, n);
        way.assign(m + 1, m);

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

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

    T get_cost() {
        T ans = 0;
        for(int j = 0; j < m; j++) {
            ans += cost[assignment[j]][j];
        }
        return ans;
    }
};

int n, m;
vector<vector<pair<int, int>>> tree_adj;
vector<tuple<int, int, int>> edges;

void read() {
    cin >> n >> m;
    edges.resize(m);
    for(auto& [u, v, w]: edges) {
        cin >> u >> v >> w;
        u--, v--;
    }
}

bool add_on_path(
    int u, int par, int tgt, int edge_tgt, vector<vector<int64_t>>& matrix
) {
    if(u == tgt) {
        return true;
    }

    for(auto [v, idx]: tree_adj[u]) {
        if(v != par && add_on_path(v, u, tgt, edge_tgt, matrix)) {
            matrix[idx][edge_tgt] = min<int64_t>(
                matrix[idx][edge_tgt],
                get<2>(edges[edge_tgt]) - get<2>(edges[idx])
            );
            return true;
        }
    }

    return false;
}

void solve() {
    // This is the inverse minimum spanning tree problem: we want to change the
    // weights as little as possible (in L1 distance) so that the first n-1
    // edges (the given tree) become a minimum spanning tree. By the MST
    // optimality condition, the tree is minimal iff for every non-tree edge e
    // its weight is at least the weight of every tree edge on the cycle it
    // closes. We may only raise tree-edge weights (potentials pot_left) and
    // lower non-tree-edge weights (potentials pot_right). For each non-tree
    // edge i and each tree edge idx on its tree path we build a cost matrix
    // matrix[idx][i] = new_w(i) - new_w(idx) lower bound, and minimizing total
    // L1 change subject to all these constraints is exactly an assignment
    // (Hungarian) problem on this matrix. The optimal potentials give the
    // reported costs: tree edges get raised by pot_left, non-tree edges get
    // lowered by pot_right.

    vector<vector<int64_t>> matrix(n - 1, vector<int64_t>(m, 0));

    tree_adj.assign(n, {});
    for(int i = 0; i < n - 1; i++) {
        auto [u, v, _] = edges[i];
        tree_adj[u].push_back({v, i});
        tree_adj[v].push_back({u, i});
    }

    for(int i = n - 1; i < m; i++) {
        auto [u, v, w] = edges[i];
        add_on_path(u, u, v, i, matrix);
    }

    auto hungarian = HungarianAlgorithm<int64_t>(matrix);
    for(int i = 0; i < n - 1; i++) {
        cout << get<2>(edges[i]) + hungarian.pot_left[i] << '\n';
    }
    for(int i = n - 1; i < m; i++) {
        cout << get<2>(edges[i]) - hungarian.pot_right[i] << '\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 with Detailed Comments

```python
import sys
sys.setrecursionlimit(10000)

def hungarian(cost):
    # cost: list of lists, size n x m, with n <= m
    n, m = len(cost), len(cost[0])
    INF = 10**15
    # potL: u[0..n], potR: v[0..m], p: match for right side, way: back-pointer
    u = [0]*(n+1)
    v = [0]*(m+1)
    p = [0]*(m+1)
    way = [0]*(m+1)
    for i in range(1, n+1):
        p[0] = i
        j0 = 0
        minv = [INF]*(m+1)
        used = [False]*(m+1)
        # Find augmenting path for row i
        while True:
            used[j0] = True
            i0 = p[j0]
            delta = INF
            j1 = 0
            # Try all columns j
            for j in range(1, m+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(m+1):
                if used[j]:
                    u[p[j]] += delta
                    v[j] -= delta
                else:
                    minv[j] -= delta
            j0 = j1
            if p[j0] == 0:
                break
        # Augment matching
        while True:
            j1 = way[j0]
            p[j0] = p[j1]
            j0 = j1
            if j0 == 0:
                break
    # p[j] = matched row for column j; duals in u[1..n], v[1..m]
    return u, v, p

def main():
    data = sys.stdin.read().split()
    it = iter(data)
    n, m = int(next(it)), int(next(it))
    edges = []
    for _ in range(m):
        u = int(next(it))-1
        v = int(next(it))-1
        w = int(next(it))
        edges.append((u, v, w))
    # Build tree adjacency (first n-1 edges)
    tree_adj = [[] for _ in range(n)]
    for i in range(n-1):
        u,v,_ = edges[i]
        tree_adj[u].append((v,i))
        tree_adj[v].append((u,i))
    # Prepare cost matrix (n-1) x m
    INF = 10**12
    matrix = [[INF]*m for _ in range(n-1)]
    # DFS to mark path
    def dfs(u, parent, target, j_edge):
        if u == target:
            return True
        for v, idx in tree_adj[u]:
            if v == parent: continue
            if dfs(v, u, target, j_edge):
                _,_,wj = edges[j_edge]
                _,_,wi = edges[idx]
                matrix[idx][j_edge] = min(matrix[idx][j_edge], wj - wi)
                return True
        return False
    # Fill matrix for each non-tree edge
    for j in range(n-1, m):
        u, v, _ = edges[j]
        dfs(u, -1, v, j)
    # Run Hungarian to get dual potentials
    u_pot, v_pot, match = hungarian(matrix)
    # Output results
    out = []
    # For tree edges, increase by u_pot[i+1]
    for i in range(n-1):
        u,v,w = edges[i]
        d = w + u_pot[i+1]
        out.append(str(d))
    # For non-tree edges, decrease by v_pot[j+1]
    for j in range(n-1, m):
        u,v,w = edges[j]
        d = w - v_pot[j+1]
        out.append(str(d))
    sys.stdout.write("\n".join(out))

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

5. Compressed Editorial

Define nonnegative increases xᵢ for each tree edge and decreases yⱼ for each non-tree edge. Constraints xᵢ + yⱼ ≤ cⱼ – cᵢ for every tree-edge i on the path of non-tree-edge j ensure the given tree stays the MST. Minimize ∑xᵢ + ∑yⱼ is a linear program whose dual is a maximum‐weight feasible potential problem on an (n–1)×M bipartite graph. Running the Hungarian algorithm on cost[i][j] = cⱼ–cᵢ when i lies on j’s path (0 otherwise, since unconstrained pairs allow the trivial bound) yields optimal duals xᵢ, yⱼ. Finally set dᵢ = cᵢ+xᵢ for tree edges and dⱼ = cⱼ–yⱼ for others.