Introduction

Bessie the cow finds herself in a peculiar situation. She’s standing in a large barn organized as an $N \times N$ grid of rooms, and she needs to visit as many rooms as possible. The challenge? Most rooms are completely dark, and Bessie can only walk into rooms that are lit.

The barn has a special lighting system with $M$ connections. Each connection is defined by two room positions: $(a, b)$ and $(x, y)$. When Bessie stands in room $(a, b)$, she can flip a switch that lights up room $(x, y)$. Initially, only the room at position $(1, 1)$ where Bessie starts is lit.

Bessie can only move to adjacent rooms (up, down, left, or right) that are already lit. The question is: how many rooms can Bessie visit by strategically using the light switches and moving through the barn?

Input Format

The input consists of:

Line 1: Two integers $N$ and $M$
- $N$: The dimension of the grid (the barn is $N \times N$)
- $M$: The number of light switch connections
Lines 2 to M+1: Four integers $a, b, x, y$ (all 1-indexed)
- Being in room $(a, b)$ allows you to light up room $(x, y)$

Example

Consider a $3 \times 3$ grid with the following connections:

N = 3, M = 3
1 1 3
1 3 1
3 3 3

This means:

From room $(1,1)$, we can light room $(1,3)$
From room $(2,1)$, we can light room $(3,1)$
From room $(2,3)$, we can light room $(3,3)$

Algorithm

The solution uses a modified Breadth-First Search (BFS) algorithm with two key events at each step:

Key Insights

Event 1 - Lighting: When we visit a room $(a, b)$, we immediately activate all light switches available from that position, lighting up all connected rooms $(x, y)$.
Event 2 - Movement: We can only move to adjacent rooms (4-directional: up, down, left, right) that are already lit.
Queue Management: We use a queue to track which rooms we can visit next. A room is added to the queue only if:
- It is lit (either initially or by a switch)
- It is adjacent to a room we’ve already visited
- We haven’t visited it yet

Explanation

graph TD
    A[Start at room 1,1] --> B[Remove room from queue]
    B --> C[Mark room as visited]
    C --> D[Light all connected rooms]
    D --> E[Check 4 adjacent neighbors]
    E --> F{Is neighbor lit and unvisited?}
    F -->|Yes| G[Add to queue]
    F -->|No| H[Skip]
    G --> I{Queue empty?}
    H --> I
    I -->|No| B
    I -->|Yes| J[Done! Count visited rooms]

Pseudocode

def count_reachable_rooms(N, connections):
    # Initialize
    lit = {(1,1)}  # Room (1,1) starts lit
    visited = set()
    queue = [(1,1)]
    
    while queue:
        # Dequeue current room
        (a, b) = queue.pop(0)
        
        # Skip if already visited
        if (a, b) in visited:
            continue
            
        # Mark as visited
        visited.add((a, b))
        
        # Event 1: Light all rooms accessible from (a,b)
        for (x, y) in connections[(a,b)]:
            lit.add((x, y))
        
        # Event 2: Check 4 neighbors
        for (na, nb) in [(a-1,b), (a+1,b), (a,b-1), (a,b+1)]:
            # If neighbor is in bounds, lit, and unvisited
            if 1 <= na <= N and 1 <= nb <= N:
                if (na, nb) in lit and (na, nb) not in visited:
                    if (na, nb) not in queue:
                        queue.append((na, nb))
    
    return len(visited)

Complexity Analysis

Time Complexity: $O(N^2 + M)$
- We visit each room at most once: $O(N^2)$
- We process each connection at most once: $O(M)$
Space Complexity: $O(N^2 + M)$
- Storage for lit rooms, visited rooms, and queue: $O(N^2)$
- Storage for connections: $O(M)$

Walkthrough Example

Let’s trace through our $3 \times 3$ example step by step:

Initial State:

Lit: $(1,1)$
Visited: ${}$
Queue: $[(1,1)]$

Step 1: Visit $(1,1)$

Light room $(1,3)$ (from connection)
Check neighbors: $(1,2)$ is not lit, $(2,1)$ is not lit
Lit: $(1,1), (1,3)$
Visited: $(1,1)$
Queue: $[]$

At this point, no adjacent rooms are lit, so we’re stuck!

Better Example: Let’s modify with connection $(1,1) \to (1,2)$:

Step 1: Visit $(1,1)$

Light room $(1,2)$
Add $(1,2)$ to queue (it’s adjacent and lit)
Visited: $(1,1)$
Queue: $[(1,2)]$

Step 2: Visit $(1,2)$

Light any connected rooms
Check $(1,3)$, $(1,1)$ (visited), $(2,2)$
Visited: $(1,1), (1,2)$
Queue: continues…

The algorithm continues until no more reachable rooms remain in the queue.

Interactive Visualization

Below is an interactive simulator where you can see the algorithm in action. Try different grid sizes and connection patterns to understand how Bessie explores the barn!

Key Takeaways

BFS with Constraints: This problem demonstrates how BFS can be modified to handle additional constraints (lighting requirements) beyond simple connectivity.
Two-Phase Processing: The algorithm elegantly separates lighting (Event 1) from movement (Event 2), making the logic clear and correct.
State Management: We track three distinct states: rooms that are lit, rooms that are visited, and rooms in our queue. Each serves a specific purpose.
Graph Construction: The connections define an implicit directed graph where edges represent light switches rather than physical paths.
Greedy Exploration: By using BFS and always lighting available rooms, we ensure we explore all reachable areas of the barn.

Practice Problems

To further practice this concept, try these related problems:

USACO Silver - Switching on the Lights: The original inspiration for this problem
Codeforces - Graph Traversal with Constraints: Similar problems involving conditional edge activation
LeetCode - Keys and Rooms: A simpler version focusing on unlocking rooms

Conclusion

The “Bessie and the Lights” problem beautifully combines graph traversal with state management and constraint satisfaction. By understanding how to separate the lighting mechanism from the movement mechanism, we can solve this efficiently using a modified BFS approach. The key insight is recognizing that we need to track both what’s reachable (lit) and what we’ve explored (visited) as two separate concepts.

Happy coding, and may your barns always be well-lit! 🐄💡