1. Abridged Problem Statement
Given a connected undirected graph of N cities (3 ≤ N ≤ 50) and M roads (1 ≤ M ≤ 500), each road i between cities a_i and b_i has a destruction cost w_i. You must choose exactly one city c to "capture" (removing it from the graph) and then pay to destroy some roads, so that the remaining graph (with c and those roads removed) becomes disconnected. Find the minimum total destruction cost, and output which roads to destroy (by their input indices).

2. Detailed Editorial

We need to pick a city c and a set of roads E′ such that removing c and E′ splits the graph into at least two components, and we want to minimize the sum of weights of E′.

Observation. After removing c, the graph G–c is still connected unless we also remove roads. To make G–c disconnected, it suffices to pick two surviving nodes u, v in (G–c) and separate them by removing a minimum-weight cut of roads. Among all choices of c and node-pairs (u,v) in G–c, we seek the global minimum cut.

The code adds a virtual city 0 adjacent to all real cities to cover the case where no city actually needs to be captured (the "captured" city is the virtual one, meaning no real vertex is removed). For each candidate captured city t from 0 to N:
1. Consider the graph G_t formed by deleting node t and all incident edges (for t=0 this removes no real edges).
2. To ensure G_t is disconnected, pick two surviving nodes s, sink among the neighbours of t and remove a minimum-weight set of roads separating s from sink in G_t. That is exactly the s–t minimum cut (max-flow) between s and sink in G_t.
3. Pick the pair (t, s, sink) with the smallest cut value.

We restrict pairs to neighbours of t because if the graph splits, the components on either side of the cut must contain different neighbours of t, so this enumeration is sufficient. Only pairs with s < sink are considered to avoid duplicates.

Implementation details:
- Use Dinic's algorithm on up to N ≤ 50 nodes.
- Remove t by simply skipping edges that touch t.
- For each edge in G_t, add two directed edges u→v and v→u with capacity = its weight.
- Compute max-flow from s to sink; that value is the min-cut.
- After the flow, find the side of the cut reachable from s via residual capacity > 0; any original edge leaving that reachable set is in the cut. Record its index.
- Track the overall minimum; finally output its total cost, the number of edges in that cut, and their (sorted) indices.

Time complexity: For each t (N+1 choices), we do at most deg(t)·(deg(t)–1)/2 flow calls. In the worst case this is O(M²/N), well within limits for N≤50, M≤500. Dinic runs very fast on small graphs.

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;
};

template<class FlowT>
class MaxFlow {
  public:
    const static FlowT finf = 1e9 + 42 + 17;

    struct Edge {
        FlowT flow, cap;
        int idx, rev, to;
        Edge() : flow(0), cap(0), idx(0), rev(0), to(0) {}
        Edge(int _to, int _rev, FlowT _flow, FlowT _cap, int _idx)
            : flow(_flow), cap(_cap), idx(_idx), rev(_rev), to(_to) {}
    };

    int n;
    vector<vector<Edge>> g;
    vector<int> dist, po;
    vector<char> used_cut;

    void init(int _n) {
        n = _n;
        g.assign(n + 1, {});
        dist.assign(n + 1, -1);
        po.assign(n + 1, 0);
        used_cut.assign(n + 1, 0);
    }

    void add_edge(int u, int v, FlowT w, int idx = -1) {
        g[u].push_back(Edge(v, g[v].size(), 0, w, idx));
        g[v].push_back(Edge(u, g[u].size() - 1, 0, 0, -1));
    }

    bool bfs(int s, int t) {
        for(int v = 0; v <= n; v++) {
            dist[v] = -1;
            po[v] = 0;
        }

        queue<int> q;
        q.push(s);
        dist[s] = 0;
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            for(Edge e: g[u]) {
                if(dist[e.to] == -1 && e.flow < e.cap) {
                    dist[e.to] = dist[u] + 1;
                    q.push(e.to);
                }
            }
        }

        return dist[t] != -1;
    }

    FlowT dfs(int u, int t, FlowT fl = finf) {
        if(u == t) {
            return fl;
        }

        for(; po[u] < (int)g[u].size(); po[u]++) {
            auto& e = g[u][po[u]];
            if(dist[e.to] == dist[u] + 1 && e.flow < e.cap) {
                FlowT f = dfs(e.to, t, min(fl, e.cap - e.flow));
                e.flow += f;
                g[e.to][e.rev].flow -= f;
                if(f > 0) {
                    return f;
                }
            }
        }

        return 0;
    }

    FlowT flow(int s, int t) {
        if(s == t) {
            return finf;
        }

        FlowT ret = 0, to_add;
        while(bfs(s, t)) {
            while((to_add = dfs(s, t))) {
                ret += to_add;
            }
        }

        return ret;
    }

    void dfs_min_cut(int u) {
        used_cut[u] = 1;
        for(auto e: g[u]) {
            if(!used_cut[e.to] && e.cap > e.flow) {
                dfs_min_cut(e.to);
            }
        }
    }

    void find_cut(int s) {
        for(int i = 0; i <= n; i++) {
            used_cut[i] = 0;
        }
        dfs_min_cut(s);
    }
};

int n, m;
vector<pair<int, pair<int, int>>> ed;
vector<vector<int>> adj;

void read() {
    cin >> n >> m;
    ed.assign(m + 1, {});
    adj.assign(n + 1, {});
    for(int i = 1; i <= m; i++) {
        cin >> ed[i].first >> ed[i].second.first >> ed[i].second.second;
        adj[ed[i].first].push_back(ed[i].second.first);
        adj[ed[i].second.first].push_back(ed[i].first);
    }
}

int answer;
vector<int> ans;
MaxFlow<int> mf;

void check(int t, int s, int sink) {
    mf.init(n);
    for(int i = 1; i <= m; i++) {
        if(ed[i].first != t && ed[i].second.first != t) {
            mf.add_edge(
                ed[i].first, ed[i].second.first, ed[i].second.second, i
            );
            mf.add_edge(
                ed[i].second.first, ed[i].first, ed[i].second.second, i
            );
        }
    }

    int curr = mf.flow(s, sink);
    if(answer == -1 || curr < answer) {
        mf.find_cut(s);
        answer = curr;
        ans.clear();
        for(int i = 0; i <= n; i++) {
            for(auto it: mf.g[i]) {
                if(mf.used_cut[i] && !mf.used_cut[it.to] && it.idx != -1) {
                    ans.push_back(it.idx);
                }
            }
        }

        sort(ans.begin(), ans.end());
        ans.erase(unique(ans.begin(), ans.end()), ans.end());
    }
}

void solve_captured(int r) {
    for(int u: adj[r]) {
        for(int v: adj[r]) {
            if(u < v && r != u && r != v) {
                check(r, u, v);
            }
        }
    }
}

void solve() {
    // The invaders capture one city t (Berland troops cannot route through it)
    // and then need at least two surviving cities disconnected from each other.
    // Cutting roads to separate a fixed pair s, sink is exactly a min cut in
    // the graph with t removed, where each undirected road becomes a pair of
    // directed edges with capacity equal to its destruction cost.
    //
    // We brute force the captured city t (including a virtual city 0 adjacent
    // to all cities, which covers the case where no city is captured) and the
    // ordered pair s < sink among neighbours of t. Both s and sink must lie on
    // either side of the eventual cut, and any separating cut must split some
    // such neighbour pair, so this enumeration is sufficient. For each choice
    // we run Dinic, take the cheapest cut overall, and recover the cut edges by
    // a reachability DFS from s over residual capacity: edges leaving the
    // reachable set with a valid input index form the plan.

    adj.resize(n + 1);
    for(int i = 1; i <= n; i++) {
        adj[0].push_back(i);
    }

    answer = -1;
    for(int i = 0; i <= n; i++) {
        solve_captured(i);
    }

    cout << answer << '\n';
    cout << ans.size() << '\n';
    cout << ans;
    if(!ans.empty()) {
        cout << '\n';
    }
}

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

    read();
    solve();

    return 0;
}
```

4. Python Solution

```python
import sys
from collections import deque

sys.setrecursionlimit(10**7)
input = sys.stdin.readline

class Dinic:
    """Dinic's algorithm for max-flow / min-cut."""
    def __init__(self, n):
        self.n = n
        self.adj = [[] for _ in range(n+1)]

    def add_edge(self, u, v, cap, idx=-1):
        """Add directed edge u->v with capacity cap and original index idx."""
        # forward edge: (to, capacity, idx, rev)
        self.adj[u].append([v, cap, idx, len(self.adj[v])])
        # backward edge: zero capacity
        self.adj[v].append([u, 0, -1, len(self.adj[u]) - 1])

    def bfs(self, s, t, level):
        """Build level graph via BFS."""
        for i in range(len(level)):
            level[i] = -1
        queue = deque([s])
        level[s] = 0
        while queue:
            u = queue.popleft()
            for v, cap, idx, rev in self.adj[u]:
                if cap > 0 and level[v] < 0:
                    level[v] = level[u] + 1
                    queue.append(v)
        return level[t] >= 0

    def dfs(self, u, t, flow, level, it):
        """Push flow along DFS in level graph."""
        if u == t:
            return flow
        for i in range(it[u], len(self.adj[u])):
            it[u] = i
            v, cap, idx, rev = self.adj[u][i]
            if cap > 0 and level[v] == level[u] + 1:
                pushed = self.dfs(v, t, min(flow, cap), level, it)
                if pushed:
                    # update residual capacities
                    self.adj[u][i][1] -= pushed
                    self.adj[v][rev][1] += pushed
                    return pushed
        return 0

    def max_flow(self, s, t):
        """Compute max flow s->t."""
        flow = 0
        level = [-1]*(self.n+1)
        while self.bfs(s, t, level):
            it = [0]*(self.n+1)
            pushed = self.dfs(s, t, 10**18, level, it)
            while pushed:
                flow += pushed
                pushed = self.dfs(s, t, 10**18, level, it)
        return flow

    def min_cut_edges(self, s):
        """After max_flow, find reachable from s in residual graph."""
        vis = [False]*(self.n+1)
        stack = [s]
        vis[s] = True
        while stack:
            u = stack.pop()
            for v, cap, idx, rev in self.adj[u]:
                # if residual cap > 0, can go
                if cap > 0 and not vis[v]:
                    vis[v] = True
                    stack.append(v)
        return vis

def main():
    n, m = map(int, input().split())
    edges = [None]*(m+1)
    cost = [0]*(m+1)
    adj = [[] for _ in range(n+1)]
    for i in range(1, m+1):
        a, b, w = map(int, input().split())
        edges[i] = (a, b)
        cost[i] = w
        adj[a].append(b)
        adj[b].append(a)

    best = None
    best_list = []

    # Try capturing each city r
    for r in range(1, n+1):
        # Only pairs of neighbors of r can become disconnected across r
        nbrs = adj[r]
        L = len(nbrs)
        for i in range(L):
            for j in range(i+1, L):
                u, v = nbrs[i], nbrs[j]
                # Build flow network on nodes 1..n excluding r
                din = Dinic(n)
                for idx in range(1, m+1):
                    a, b = edges[idx]
                    if a == r or b == r:
                        continue
                    # Undirected -> two directed edges
                    din.add_edge(a, b, cost[idx], idx)
                    din.add_edge(b, a, cost[idx], idx)
                # Compute min-cut between u and v
                f = din.max_flow(u, v)
                if best is None or f < best:
                    # Identify which original edges are in the cut
                    reachable = din.min_cut_edges(u)
                    cut_edges = set()
                    for x in range(1, n+1):
                        if not reachable[x]: continue
                        for to, cap, idx, rev in din.adj[x]:
                            if not reachable[to] and idx != -1:
                                cut_edges.add(idx)
                    cut_list = sorted(cut_edges)
                    best = f
                    best_list = cut_list

    # Output
    print(best)
    print(len(best_list))
    if best_list:
        print(*best_list)

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

5. Compressed Editorial

- We must capture a city t and remove roads to disconnect G–t.
- That is equivalent to finding an s–t minimum cut in G–t for some two survivors s, sink.
- A virtual city 0 adjacent to all nodes is added to cover the case where no real city needs to be captured.
- Try every t∈[0..N], and among its neighbours s < sink compute min-cut(s, sink) on G with t removed.
- Track the global minimum and record which edges cross that cut via the residual graph.
- Use Dinic's algorithm on N≤50, M≤500.
