Overview

This problem asks you to determine whether all evacuees in a building can reach the emergency exit while avoiding rooms on fire, and if possible, find the minimum number of turns for everyone to complete evacuation.

Analysis

Key Insight: Evacuees Do Not Affect Each Other

The problem statement explicitly states that “there is no upper limit on the number of people in corridors or rooms” and “each evacuee’s actions do not affect others.” This is a crucial condition, meaning that each evacuee can independently take their shortest path to evacuate.

In other words, the optimal action for evacuee \(i\) is to follow the shortest path to room \(T\) while avoiding rooms on fire.

Deriving the Answer

Since each evacuee can move independently: - Minimum number of turns for evacuee \(i\) to complete evacuation = shortest distance from \(S_i\) to \(T\) without passing through rooms on fire - Minimum number of turns for everyone to complete evacuation = maximum value among all evacuees

If even one evacuee cannot reach \(T\) while avoiding the fire, the answer is \(-1\).

Why Passing Through Rooms on Fire Is Impossible

In one turn, you can move to at most one adjacent room. The moment you move into a room on fire, the elimination check causes you to be eliminated, making it impossible to pass through a room on fire as a waypoint. Therefore, we simply need to find the shortest distance on a graph with rooms on fire completely removed.

Algorithm

1. If any evacuee’s initial position is a room on fire, immediately output -1
2. Starting from the emergency exit \(T\), perform BFS (Breadth-First Search) on the graph with rooms on fire excluded to find the shortest distance to each room
3. For each evacuee’s initial position \(S_i\):
   • If \(S_i = T\), evacuation is complete at turn \(0\) (distance \(0\))
   • If \(S_i\) is unreachable (distance is undefined), output -1
   • Otherwise, record the distance
4. The maximum distance among all evacuees is the answer

Concrete Example

If there are 5 rooms, \(T = 1\), and a fire has occurred in room 3:

1 - 2 - 3 - 4 - 5

Since room 3 is on fire, rooms 4 and 5 cannot reach room 1 (cannot pass through room 3). If there are evacuees in rooms 4 or 5, the answer is -1.

Complexity

• Time complexity: \(O(N + M + K)\)
  • BFS: \(O(N + M)\)
  • Checking distance for each evacuee: \(O(K)\)
• Space complexity: \(O(N + M)\)
  • Adjacency list: \(O(N + M)\)
  • Distance array: \(O(N)\)

Implementation Notes

• By setting the BFS starting point to \(T\) (the exit), we can compute the distance from every room to \(T\) with a single BFS (since the graph is undirected, distance from \(T\) = distance to \(T\))

• Rooms on fire are excluded from BFS visitation (the v not in fire condition)

• By managing rooms on fire in a set, we can perform lookups in \(O(1)\)

• When \(Q = 0\) (no fire), care is needed for empty line input, but reading with split() handles this without issues

  Source Code

import sys
from collections import deque

def main():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    M = int(input_data[idx]); idx += 1
    K = int(input_data[idx]); idx += 1
    Q = int(input_data[idx]); idx += 1
    T = int(input_data[idx]); idx += 1
    
    adj = [[] for _ in range(N + 1)]
    for _ in range(M):
        u = int(input_data[idx]); idx += 1
        v = int(input_data[idx]); idx += 1
        adj[u].append(v)
        adj[v].append(u)
    
    S = []
    for i in range(K):
        S.append(int(input_data[idx])); idx += 1
    
    fire = set()
    for i in range(Q):
        fire.add(int(input_data[idx])); idx += 1
    
    # Check if any evacuee starts on fire
    for s in S:
        if s in fire:
            print(-1)
            return
    
    # BFS from T, avoiding fire rooms
    dist = [-1] * (N + 1)
    dist[T] = 0
    queue = deque()
    queue.append(T)
    
    while queue:
        u = queue.popleft()
        for v in adj[u]:
            if dist[v] == -1 and v not in fire:
                dist[v] = dist[u] + 1
                queue.append(v)
    
    ans = 0
    for s in S:
        if s == T:
            continue
        if dist[s] == -1:
            print(-1)
            return
        if dist[s] > ans:
            ans = dist[s]
    
    print(ans)

main()

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