Answer:
from collections import deque
def find_most_accessible_buildings(n, m, routes):
# Create an adjacency list to represent the graph
graph = [[] for _ in range(n)]
for u, v in routes:
graph[u].append(v)
# Initialize the lists to keep track of the distances and counts
distances = [-1] * n
counts = [0] * n
# Perform BFS starting from each building
for start in range(n):
visited = [False] * n
queue = deque()
queue.append((start, 0))
visited[start] = True
distances[start] = 0
while queue:
current, distance = queue.popleft()
counts[distance] += 1
for neighbor in graph[current]:
if not visited[neighbor]:
visited[neighbor] = True
queue.append((neighbor, distance + 1))
distances[neighbor] = distance + 1
# Find the maximum count and the buildings with that count
max_count = max(counts)
accessible_buildings = [i for i, count in enumerate(counts) if count == max_count]
return max_count, accessible_buildings
# Read the input
n, m = map(int, input().split())
routes = [tuple(map(int, input().split())) for _ in range(m)]
# Call the function to find the most accessible buildings
max_count, accessible_buildings = find_most_accessible_buildings(n, m, routes)
# Print the output
print(max_count)
print(' '.join(map(str, accessible_buildings)))
Step-by-step explanation:
To solve this problem, we can use graph traversal algorithms like Breadth-First Search (BFS) to find the number of buildings accessible from each building. Here's a Python implementation for this problem:
```python
from collections import deque
def find_most_accessible_buildings(n, m, routes):
# Create an adjacency list to represent the graph
graph = [[] for _ in range(n)]
for u, v in routes:
graph[u].append(v)
# Initialize the lists to keep track of the distances and counts
distances = [-1] * n
counts = [0] * n
# Perform BFS starting from each building
for start in range(n):
visited = [False] * n
queue = deque()
queue.append((start, 0))
visited[start] = True
distances[start] = 0
while queue:
current, distance = queue.popleft()
counts[distance] += 1
for neighbor in graph[current]:
if not visited[neighbor]:
visited[neighbor] = True
queue.append((neighbor, distance + 1))
distances[neighbor] = distance + 1
# Find the maximum count and the buildings with that count
max_count = max(counts)
accessible_buildings = [i for i, count in enumerate(counts) if count == max_count]
return max_count, accessible_buildings
# Read the input
n, m = map(int, input().split())
routes = [tuple(map(int, input().split())) for _ in range(m)]
# Call the function to find the most accessible buildings
max_count, accessible_buildings = find_most_accessible_buildings(n, m, routes)
# Print the output
print(max_count)
print(' '.join(map(str, accessible_buildings)))
```
In this implementation, the function `find_most_accessible_buildings` takes the number of buildings `n`, the number of bus routes `m`, and a list of `routes` as input. It creates an adjacency list `graph` to represent the connections between buildings. It then performs a BFS starting from each building to find the distances and counts of accessible buildings. Finally, it finds the maximum count and the buildings with that count.
In the main part of the code, it reads the input, calls the `find_most_accessible_buildings` function, and prints the output as specified in the problem statement.
You can run this code and provide the input as described in the problem to get the corresponding output.