1. Abridged Problem Statement  
Given an undirected graph of n intersections (numbered 1..n) where each pair (i, j) has a nonnegative road length lij (0 means no road), and a sequence of k positive integers r₁…rₖ (the lengths of roads actually traveled, in order), find all intersections you can end up at if you start from intersection 1 and follow a path whose edge‐lengths exactly match r₁…rₖ. Nodes and edges may be revisited. Output the count of such end‐nodes and their sorted list (1‐based). If none, output 0.

2. Detailed Editorial  

We need to check all walks of length k (in terms of number of edges) starting at node 1, such that the sequence of edge lengths exactly matches the given sequence r[0…k−1]. Because n, k ≤ 200, the total number of walks is exponential if explored naively; instead, we do a BFS/DP on a state space of size n×(k+1):

  State = (current_node, position_in_sequence)  
    – current_node ∈ [0..n−1]  
    – position_in_sequence ∈ [0..k]  

Meaning: we have just matched r[0…position−1] and are now at current_node. From any state (u, p) with p<k, we can move to any neighbor v of u whose edge‐length equals r[p], reaching state (v, p+1). When p==k, we have matched the entire sequence and u is one possible secret‐place.  

Use a boolean visited[n][k+1] to avoid re‐processing the same state twice. Initialize a queue with (0, 0) (since we start at node 1, 0‐indexed, having matched zero elements). Standard BFS (or simple queue‐driven DP) explores all reachable states. Whenever we dequeue (u, k), record u in an answer set.  

Time complexity: For each of up to n×k states, we scan O(n) neighbors → O(n²·k) = max 200²·200 = 8×10⁶ operations, easily within 0.5 s in C++.

3. C++ Solution with Detailed Comments  

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

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

    int n;
    cin >> n;  
    // Read adjacency matrix of road lengths
    vector<vector<int>> graph(n, vector<int>(n));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> graph[i][j];
        }
    }

    int k;
    cin >> k;  
    // Read the sequence of road lengths traveled
    vector<int> r(k);
    for (int i = 0; i < k; i++) {
        cin >> r[i];
    }

    // visited[u][p] = true if we've reached node u after matching p edges
    vector<vector<bool>> visited(n, vector<bool>(k + 1, false));
    // Standard BFS queue of pairs (current_node, matched_length)
    deque<pair<int,int>> q;

    // Start from node 0 (intersection 1), with 0 edges matched
    visited[0][0] = true;
    q.emplace_back(0, 0);

    // Use a set to collect all end-nodes reached after k steps
    set<int> possible;

    while (!q.empty()) {
        auto [u, pos] = q.front();
        q.pop_front();

        // If we've matched all k edges, record this node as possible end
        if (pos == k) {
            possible.insert(u);
            continue;
        }

        // Otherwise try all neighbors v; if graph[u][v] == r[pos], we can move
        for (int v = 0; v < n; v++) {
            if (graph[u][v] == r[pos] && !visited[v][pos + 1]) {
                visited[v][pos + 1] = true;
                q.emplace_back(v, pos + 1);
            }
        }
    }

    // Output result
    if (possible.empty()) {
        cout << 0 << "\n";
    } else {
        cout << possible.size() << "\n";
        // Output in 1-based indexing
        bool first = true;
        for (int u : possible) {
            if (!first) cout << ' ';
            first = false;
            cout << (u + 1);
        }
        cout << "\n";
    }

    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
from collections import deque

def solve():
    # Number of intersections
    n = int(input())
    # Read adjacency matrix: graph[i][j] = length between i+1 and j+1, or 0 if no road
    graph = [list(map(int, input().split())) for _ in range(n)]
    
    # Number of roads traveled
    k = int(input())
    # The exact sequence of road lengths observed
    path = list(map(int, input().split()))
    
    # BFS queue of states: (current_node, how_many_roads_matched)
    queue = deque([(0, 0)])  # start at node 0, matched 0 roads
    # visited[node][pos] marks we've reached `node` after matching pos edges
    visited = [[False] * (k + 1) for _ in range(n)]
    visited[0][0] = True
    
    possible = set()  # all end-nodes once pos == k

    while queue:
        node, pos = queue.popleft()
        
        # If we've matched the full sequence, record this node
        if pos == k:
            possible.add(node)
            continue
        
        # Try every neighbor for the next step
        needed_length = path[pos]
        for nxt in range(n):
            # If there's a road of the needed length and we haven't visited (nxt,pos+1)
            if graph[node][nxt] == needed_length and not visited[nxt][pos + 1]:
                visited[nxt][pos + 1] = True
                queue.append((nxt, pos + 1))
    
    # Prepare output
    possible = sorted(possible)
    if not possible:
        print(0)
    else:
        # Number of possible secret places
        print(len(possible))
        # Convert to 1-based indexing
        print(*[x + 1 for x in possible])

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

5. Compressed Editorial  
Perform a BFS/DP on states (node, steps_matched), transitions follow edges whose length equals the next required value in the observed sequence. Mark visited states to avoid repetition. Collect all nodes reached when steps_matched == k. Complexity O(n²·k), acceptable for n,k ≤ 200.