1. Abridged Problem Statement  
Given a directed graph of n cities and m one-way roads (edges), you may choose at most one of these roads and make it two-way (i.e. add its reverse). Define the “degree of development” as the size of the largest strongly connected subset of cities (i.e. largest strongly connected component).    
• Compute the maximum degree of development w achievable by upgrading at most one road.  
• List all road indices which, when made two-way, achieve this maximum w.  

2. Detailed Editorial  

Overview  
We must consider, for each original edge e = (u→v), what happens if we add the reverse edge v→u. That additional edge can merge multiple strongly connected components (SCCs) in the condensation DAG. We need to compute the resulting largest SCC size for each choice, pick the maximum size w, and report all edges e achieving w.

Step A: Compute SCCs of the original graph  
– Run Kosaraju (or Tarjan) to label each vertex with its SCC index comp[v], total C SCCs.  
– Record comp_size[c] = number of original vertices in SCC c.  
– Build the condensation DAG of these C components: for each original edge (u→v) if comp[u]≠comp[v], add a DAG edge comp[u]→comp[v]. Remove duplicates.

Step B: Topological order and reachability closure  
– Compute a topological order topo of the DAG.  
– Build forward reachability closure[f][g]: bitset of components reachable from f (including f).  
  Process in reverse topo: for u in reverse(topo),  
    closure[u] = union over closure[v] for all v in adj[u], plus bit u itself.  
– Similarly build reverse reachability closure_rev[g][f]: those that can reach g. This is just the forward closure on the reversed DAG.

Step C: Evaluate each edge  
For each original road i: u→v  
  let cu = comp[u], cv = comp[v].  
  If cu==cv, upgrading does nothing; the largest SCC remains the size comp_size[cu].  
  Else:  
    Merged components = { c : closure[cu][c]=1 AND closure_rev[cv][c]=1 }.  
    merged_size = sum(comp_size[c] for c in Merged components)  
    best_other = maximum comp_size[d] among d not in Merged components  
    current_max = max(merged_size, best_other)  
  Track the global maximum w and collect all edges i for which current_max = w.

Complexities  
– SCC (Kosaraju): O(n+m)  
– Building DAG & topo: O(C + #dag_edges) ≤ O(n + m)  
– Computing closure with bitsets: O(C²/w)  
– Loop over m edges each in O(C + C) = O(m·C) ≤ 2·10^7 in worst case. Acceptable for n≤1000, m≤20000.

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Class to find strongly connected components (Kosaraju)
class StronglyConnectedComponents {
  private:
    vector<bool> visited;              // visited flag for the first DFS

    // First DFS: build a reverse topological order in top_sort
    void dfs1(int u) {
        visited[u] = true;
        for (int v : adj[u]) {
            if (!visited[v])
                dfs1(v);
        }
        top_sort.push_back(u);  // post-order
    }

    // Second DFS on the reversed graph: assign component IDs
    void dfs2(int u) {
        for (int v : radj[u]) {
            if (comp[v] == -1) {
                comp[v] = comp[u];  // same component
                dfs2(v);
            }
        }
    }

  public:
    int n;                           // number of vertices
    vector<vector<int>> adj, radj;   // adjacency and reverse adjacency
    vector<int> comp;                // component assignment for each vertex
    vector<int> comp_ids;            // list of component IDs found
    vector<int> top_sort;            // vertices in postorder of dfs1

    StronglyConnectedComponents() {}
    StronglyConnectedComponents(int _n) { init(_n); }

    // Add directed edge u->v
    void add_edge(int u, int v) {
        adj[u].push_back(v);
        radj[v].push_back(u);
    }

    // Initialize for a new graph of size _n
    void init(int _n) {
        n = _n;
        comp_ids.clear();
        top_sort.clear();
        adj.assign(n, {});
        radj.assign(n, {});
    }

    // Main routine: find all SCCs
    void find_components() {
        comp.assign(n, -1);
        visited.assign(n, false);
        // 1) Order vertices by finish time
        for (int i = 0; i < n; i++) {
            if (!visited[i]) dfs1(i);
        }
        // 2) Process in reverse postorder on the reversed graph
        reverse(top_sort.begin(), top_sort.end());
        for (int u : top_sort) {
            if (comp[u] == -1) {
                comp[u] = comp_ids.size();     // new component ID
                comp_ids.push_back(comp[u]);
                dfs2(u);
            }
        }
    }
};

int n, m;
vector<pair<int,int>> edges;   // original edges (0-based)
StronglyConnectedComponents SCC;

// Read input
void read() {
    cin >> n >> m;
    edges.resize(m);
    SCC.init(n);
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        u--, v--;
        edges[i] = {u, v};
        SCC.add_edge(u, v);
    }
}

void solve() {
    // 1) Compute SCCs
    SCC.find_components();
    int C = SCC.comp_ids.size();  // number of components

    // 2) Build condensation DAG
    vector<vector<bool>> used(C, vector<bool>(C, false));
    vector<vector<int>> cadj(C);
    vector<int> indegree(C, 0);
    for (auto &e : edges) {
        int cu = SCC.comp[e.first], cv = SCC.comp[e.second];
        if (cu != cv && !used[cu][cv]) {
            used[cu][cv] = true;
            cadj[cu].push_back(cv);
            indegree[cv]++;
        }
    }

    // 3) Record sizes of each component
    vector<int> comp_size(C, 0);
    for (int v = 0; v < n; v++)
        comp_size[SCC.comp[v]]++;

    // 4) Topological sort of the DAG
    queue<int> q;
    for (int i = 0; i < C; i++)
        if (indegree[i] == 0) q.push(i);

    vector<int> topo;
    while (!q.empty()) {
        int u = q.front(); q.pop();
        topo.push_back(u);
        for (int w : cadj[u]) {
            if (--indegree[w] == 0)
                q.push(w);
        }
    }

    // 5) Compute forward reachability closure with bitsets
    vector<bitset<1024>> closure(C);
    for (int i = 0; i < C; i++)
        closure[i].set(i);   // each reaches itself
    // process in reverse topo so that children are filled first
    for (int i = C - 1; i >= 0; i--) {
        int u = topo[i];
        for (int v : cadj[u])
            closure[u] |= closure[v];
    }

    // 6) Also compute reverse closure (who can reach you)
    vector<vector<int>> rcadj(C);
    for (int u = 0; u < C; u++)
        for (int v : cadj[u])
            rcadj[v].push_back(u);

    vector<bitset<1024>> rclosure(C);
    for (int i = 0; i < C; i++)
        rclosure[i].set(i);
    for (int i = C - 1; i >= 0; i--) {
        int u = topo[i];
        for (int v : rcadj[u])
            rclosure[u] |= rclosure[v];
    }

    int best = 0;
    vector<int> ans_edges;

    // 7) Try each original edge
    if (m == 0) {
        // Special case: no edges, best we can do is pick any single city
        best = 1;
    } else {
        for (int i = 0; i < m; i++) {
            auto [u, v] = edges[i];
            int cu = SCC.comp[u], cv = SCC.comp[v];

            int current_max;
            if (cu == cv) {
                // No change
                current_max = comp_size[cu];
            } else {
                // Components that become one big SCC
                bitset<1024> merged = closure[cu] & rclosure[cv];
                int merged_size = 0, other_max = 0;
                for (int c = 0; c < C; c++) {
                    if (merged.test(c))
                        merged_size += comp_size[c];
                    else
                        other_max = max(other_max, comp_size[c]);
                }
                current_max = max(merged_size, other_max);
            }
            // Track the global maximum
            if (current_max > best) {
                best = current_max;
                ans_edges.clear();
                ans_edges.push_back(i+1);
            } else if (current_max == best) {
                ans_edges.push_back(i+1);
            }
        }
    }

    // 8) Output
    cout << best << "\n";
    cout << ans_edges.size() << "\n";
    for (int i = 0; i < (int)ans_edges.size(); i++) {
        if (i) cout << " ";
        cout << ans_edges[i];
    }
    if (!ans_edges.empty()) cout << "\n";
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
sys.setrecursionlimit(10**7)

def readints():
    return map(int, sys.stdin.readline().split())

# 1) Read input
n, m = map(int, sys.stdin.readline().split())
edges = []
g = [[] for _ in range(n)]
rg = [[] for _ in range(n)]
for _ in range(m):
    u, v = map(int, sys.stdin.readline().split())
    u -= 1; v -= 1
    edges.append((u, v))
    g[u].append(v)
    rg[v].append(u)

# 2) Kosaraju to find SCCs
visited = [False]*n
order = []
def dfs1(u):
    visited[u] = True
    for w in g[u]:
        if not visited[w]:
            dfs1(w)
    order.append(u)

for i in range(n):
    if not visited[i]:
        dfs1(i)

comp = [-1]*n
cid = 0
def dfs2(u, cid):
    comp[u] = cid
    for w in rg[u]:
        if comp[w] == -1:
            dfs2(w, cid)

# process in reverse finishing order
for u in reversed(order):
    if comp[u] == -1:
        dfs2(u, cid)
        cid += 1

C = cid  # number of components

# 3) comp_size and build DAG of components
comp_size = [0]*C
for v in range(n):
    comp_size[comp[v]] += 1

cadj = [set() for _ in range(C)]
rcadj = [set() for _ in range(C)]
for u, v in edges:
    cu, cv = comp[u], comp[v]
    if cu != cv:
        cadj[cu].add(cv)
        rcadj[cv].add(cu)

# 4) Topological sort of the DAG (Kahn’s algorithm)
indeg = [0]*C
for u in range(C):
    for v in cadj[u]:
        indeg[v] += 1

from collections import deque
q = deque(i for i in range(C) if indeg[i] == 0)
topo = []
while q:
    u = q.popleft()
    topo.append(u)
    for v in cadj[u]:
        indeg[v] -= 1
        if indeg[v] == 0:
            q.append(v)

# 5) Forward and reverse reachability with Python ints as bitsets
closure = [0]*C
rclosure = [0]*C
# each comp can reach itself
for i in range(C):
    closure[i] = 1 << i
    rclosure[i] = 1 << i

# forward closure: process reverse topo
for u in reversed(topo):
    for v in cadj[u]:
        closure[u] |= closure[v]

# reverse closure: on reversed DAG
for u in reversed(topo):
    for v in rcadj[u]:
        rclosure[u] |= rclosure[v]

# 6) Evaluate each edge
best = 0
ans = []

if m == 0:
    # no edges: best is any single node
    best = 1
else:
    for idx, (u, v) in enumerate(edges, start=1):
        cu, cv = comp[u], comp[v]
        if cu == cv:
            cur = comp_size[cu]
        else:
            # components that merge are those c with
            # bit c set in both closure[cu] and rclosure[cv]
            merged_mask = closure[cu] & rclosure[cv]
            merged_size = 0
            other_max = 0
            for c in range(C):
                if (merged_mask >> c) & 1:
                    merged_size += comp_size[c]
                else:
                    other_max = max(other_max, comp_size[c])
            cur = max(merged_size, other_max)
        if cur > best:
            best = cur
            ans = [idx]
        elif cur == best:
            ans.append(idx)

# 7) Output result
print(best)
print(len(ans))
if ans:
    print(*ans)
```

5. Compressed Editorial  
– Find SCCs, form condensation DAG.  
– Compute forward reachability closure and reverse closure on the DAG via bitsets in topological order.  
– For each original edge (u→v), if it links different components cu→cv, merging occurs on all c where cu→…→c→…→cv. Sum their sizes, compare with the largest unaffected component. If cu=cv, no change.  
– Track the maximum resulting SCC size w and list all edges achieving w.