1. Abridged Problem Statement
Given a directed tree of N spies (nodes) with unique paths from every spy to the root (spy 1). Each of the N−1 edges is either "protected" or "almost protected." A path is safe if at least half of its edges are protected. You may upgrade some almost‐protected edges to protected. Find a minimum set of edges to upgrade so that every path from any spy to spy 1 is safe, and output their input indices.

2. Detailed Editorial

• Root the tree at node 1 and do a DFS, keeping track of the current depth d (number of edges from root) and the set S of "almost‐protected" edges on the current root‐to‐u path.
• At any node u, there are d total edges on its path, of which |S| are almost. The safety condition is: number of protected ≥ number of almost ⇒ d−|S| ≥ |S| ⇒ 2|S| ≤ d.
• If 2|S| > d at node u, we must immediately upgrade one of the almost‐protected edges on the path. To maximize the benefit (so that future deeper nodes also have fewer almost edges), we choose the one closest to the root—that is, the almost edge with minimum depth. Remove it from S (meaning we treat it as protected hereafter), and record its index in the answer list.
• Continue DFS, inserting each almost edge (keyed by its depth and its input index) when descending, and removing it when backtracking—unless it was already removed by an upgrade.
• This greedy is optimal because we only upgrade when strictly necessary, and each upgrade is placed to cover as many future violations as possible. Each edge is inserted/removed at most once, and each upgrade uses a set operation, so total O(N log N).

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;
};

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

void read() {
    cin >> n;
    adj.assign(n, {});
    for(int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        u--, v--;

        string type;
        getline(cin, type);
        if(type[1] == 'a') {
            adj[u].emplace_back(v, i);
            adj[v].emplace_back(u, i);
        } else if(type[1] == 'p') {
            adj[u].emplace_back(v, -1);
            adj[v].emplace_back(u, -1);
        }
    }
}

void solve() {
    // All channels point towards Kat (node 1), forming a tree rooted there.
    // For a node at depth d (the path to Kat has d edges) we need at least
    // ceil(d / 2) of those edges protected, i.e. at most floor(d / 2) may
    // stay almost protected. We DFS from the root and maintain the set
    // almost_protected of the almost-protected edges on the current
    // root-to-node path, keyed by (depth_of_edge, edge_index).
    //
    // When entering a node at depth d, if the number of almost-protected
    // edges on the path exceeds floor(d / 2), we must upgrade one of them.
    // We greedily upgrade the shallowest such edge (smallest depth), since
    // it is shared by the most descendants and therefore helps the most
    // subtrees at once; that edge is removed from the set and recorded in
    // ans. Backtracking removes each edge we added on the way down.

    vector<int> ans;
    set<pair<int, int>> almost_protected;

    function<void(int, int, int)> dfs = [&](int u, int p, int depth) {
        if(almost_protected.size() * 2 > depth) {
            auto lowest = almost_protected.begin();
            ans.push_back(lowest->second);
            almost_protected.erase(lowest);
        }

        for(auto [v, i]: adj[u]) {
            if(v == p) {
                continue;
            }
            if(i != -1) {
                almost_protected.insert({depth, i});
            }

            dfs(v, u, depth + 1);
            if(i != -1 && almost_protected.count({depth, i})) {
                almost_protected.erase({depth, i});
            }
        }
    };

    dfs(0, -1, 0);

    cout << ans.size() << endl;
    for(int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " \n"[i + 1 == ans.size()];
    }
}

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
```python
import sys
sys.setrecursionlimit(300000)
input = sys.stdin.readline

n = int(input())
adj = [[] for _ in range(n)]
# Read edges; index edges 1..n-1
for idx in range(1, n):
    u, v, *rest = input().split()
    u, v = int(u)-1, int(v)-1
    # rest is ['almost', 'protected'] or ['protected']
    if rest[0][0] == 'a':  # "almost"
        # mark almost edges with their input index
        adj[u].append((v, idx))
        adj[v].append((u, idx))
    else:
        # protected edges carry -1
        adj[u].append((v, -1))
        adj[v].append((u, -1))

import heapq
alive = set()      # currently active almost‐protected edges: (depth, idx)
heap = []          # min‐heap of (depth, idx)
ans = []

def dfs(u, p, depth):
    # If too many almost edges → violation
    if len(alive)*2 > depth:
        # Extract the earliest (smallest depth) almost edge
        while True:
            d, tag = heap[0]
            if (d, tag) in alive:
                break
            heapq.heappop(heap)  # discard stale entries
        d, tag = heapq.heappop(heap)
        alive.remove((d, tag))
        ans.append(tag)  # record its index

    for v, eidx in adj[u]:
        if v == p:
            continue
        if eidx != -1:
            alive.add((depth, eidx))
            heapq.heappush(heap, (depth, eidx))
        # Recurse
        dfs(v, u, depth+1)
        # On backtrack remove if still present
        if eidx != -1 and (depth, eidx) in alive:
            alive.remove((depth, eidx))

dfs(0, -1, 0)

# Output result
print(len(ans))
print(*ans)
```

5. Compressed Editorial
Perform a DFS from the root, tracking the set S of almost-protected edges on the current path keyed by (depth, index). At each node of depth d, if 2|S|>d, greedily convert (remove from S and record) the almost-protected edge of minimum depth. This ensures all root‐to‐node paths satisfy protected≥almost, uses the fewest upgrades, and runs in O(N log N).
