Overview

This problem can be rephrased as finding the length of the shortest path from intersection \(N\) to intersection \(1\) without passing through intersections with barricades.

The answer is the shortest distance from \(N\) to \(1\) on the graph excluding the barricaded intersections.

Analysis

The kitten can only move to intersections where a signpost has been placed.

In other words, for the kitten to move to a certain intersection, a signpost must be placed at that intersection.

For example, suppose we want to guide the kitten along the following path:

\[ N \rightarrow a \rightarrow b \rightarrow 1 \]

In this case, signposts must be placed at \(a, b,\) and \(1\).

Therefore, the number of signposts required for this path is \(3\), which is the same as the number of roads traversed.

In particular, a signpost is also required at intersection \(1\) to move to intersection \(1\).

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The key observations are as follows:

Any successful guidance requires at least as many signposts as the number of moves

Suppose the kitten actually moves as follows:

\[ N = v_0 \rightarrow v_1 \rightarrow v_2 \rightarrow \cdots \rightarrow v_L = 1 \]

For each \(v_i \ (i \geq 1)\), a signpost is required to move to it.

Also, since signposts at intersections other than \(1\) fall down after the kitten arrives, the kitten cannot move to the same intersection multiple times.

Thus, at least \(L\) signposts are required.

Since this sequence of moves is a path that only passes through non-barricaded intersections, if we let \(D\) be the shortest distance from \(N\) to \(1\):

\[ D \leq L \]

Therefore, the number of signposts required is at least \(D\).

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Placing signposts along the shortest path is sufficient

Conversely, let the shortest path avoiding the barricades be:

\[ N = v_0 \rightarrow v_1 \rightarrow \cdots \rightarrow v_D = 1 \]

If we place signposts at \(v_1, v_2, \ldots, v_D\), the number of signposts used is \(D\).

The kitten moves to the adjacent intersection with the smallest index among those that have active signposts.

At first glance, it might seem that the kitten could go in an unintended direction depending on the indices of the intersections.

However, on the shortest path, there is never a direct road from the current intersection \(v_i\) to a future intersection on the path \(v_j \ (j > i + 1)\).

If such a road existed, we could replace the subsegment

\[ v_i \rightarrow v_{i+1} \rightarrow \cdots \rightarrow v_j \]

with a direct edge

\[ v_i \rightarrow v_j \]

to create a shorter path, contradicting the assumption that this is the shortest path.

Therefore, the only active signpost reachable from the current location on the shortest path is essentially the next intersection \(v_{i+1}\).

Since the signposts of the previously visited intersections have already fallen down, the kitten will follow the shortest path and eventually arrive at intersection \(1\).

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From the above, the answer is simply:

The shortest distance from \(N\) to \(1\), assuming we cannot pass through the barricaded intersections.

A naive brute-force search over which intersections to place signposts would have \(2^N\) choices, which is far too slow to pass within the time limit.

Instead, we can find the shortest distance in the graph using BFS.

Algorithm

The graph is an unweighted graph where all roads have the same length.

Therefore, the shortest distance can be found using BFS (Breadth-First Search).

The procedure is as follows:

1. Represent the graph using an adjacency list.
2. Record the barricaded intersections in a blocked array.
3. Start BFS from intersection \(N\).
4. If an adjacent intersection is barricaded, ignore it.
5. If the adjacent intersection has not been visited yet, update its distance and push it to the queue.
6. Finally, output dist[1].
   • If it is unreachable, dist[1] remains -1, so output -1.

Complexity

• Time Complexity: \(O(N + M)\)
• Space Complexity: \(O(N + M)\)

Implementation Points

• Since signposts cannot be placed at barricaded intersections, we should avoid moving to those intersections during the BFS.

• It is guaranteed that there are no barricades at intersection \(1\) and intersection \(N\).

• We treat dist[v] = -1 as “unvisited”.

• The starting point of the BFS is intersection \(N\), which is the kitten’s starting location.

  Source Code

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

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

    int N, M, K;
    cin >> N >> M >> K;

    vector<vector<int>> G(N + 1);
    for (int i = 0; i < M; i++) {
        int u, v;
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }

    vector<bool> blocked(N + 1, false);
    for (int i = 0; i < K; i++) {
        int b;
        cin >> b;
        blocked[b] = true;
    }

    vector<int> dist(N + 1, -1);
    queue<int> q;

    dist[N] = 0;
    q.push(N);

    while (!q.empty()) {
        int v = q.front();
        q.pop();

        for (int to : G[v]) {
            if (blocked[to]) continue;
            if (dist[to] != -1) continue;

            dist[to] = dist[v] + 1;
            q.push(to);
        }
    }

    cout << dist[1] << '\n';

    return 0;
}

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This editorial was generated by gpt-5.5-high.