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 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

```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, k;
vector<vector<int>> graph;
vector<int> path;

void read() {
    cin >> n;
    graph.assign(n, vector<int>(n));
    cin >> graph;
    cin >> k;
    path.resize(k);
    cin >> path;
}

void solve() {
    // BFS over states (node, pos): we are at intersection node having
    // matched the first pos road lengths of the route. From (node, pos) we
    // step to any next_node whose connecting road length graph[node][next]
    // equals path[pos], advancing to (next, pos + 1). visited[node][pos]
    // prevents revisiting a state. Every state reaching pos == k marks node
    // as a possible secret place. We collect those nodes (1-indexed),
    // sorted ascending.

    deque<pair<int, int>> queue = {{0, 0}};
    vector<vector<bool>> visited(n, vector<bool>(k + 1, false));
    visited[0][0] = true;
    set<int> possible;

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

        if(pos == k) {
            possible.insert(node);
            continue;
        }

        for(int next_node = 0; next_node < n; next_node++) {
            if(graph[node][next_node] == path[pos] &&
               !visited[next_node][pos + 1]) {
                visited[next_node][pos + 1] = true;
                queue.push_back({next_node, pos + 1});
            }
        }
    }

    cout << possible.size() << '\n';
    if(!possible.empty()) {
        bool first = true;
        for(int x: possible) {
            cout << (first ? "" : " ") << x + 1;
            first = false;
        }
        cout << '\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

```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.
