Overview

This is a shortest path problem where you need to find the minimum number of moves to reach the goal from the start on a slippery ice rink. There is a special movement rule: once you start sliding, you cannot stop until you hit a wall or stop just before a pillar.

Analysis

Problem Characteristics

The key point of this problem is the movement rule: “once you start sliding, you cannot stop.”

For example, on a \(5 \times 5\) rink, if you slide right from \((1, 1)\), you will move all the way to \((1, 5)\) (the right wall) if there are no pillars in the way. If there is a pillar at \((1, 3)\), you will stop at \((1, 2)\) (just before the pillar).

Why BFS Works

• Each move has a uniform cost of “1”
• We want to find the minimum number of moves

For such problems, Breadth-First Search (BFS) is optimal. BFS guarantees that when you first reach a state, it is at the shortest distance.

State Definition

The state can be simply represented by “the current coordinates \((r, c)\)”. Since there is no need to reach the same cell multiple times with the same number of moves, we can track visited states.

Algorithm

1. Initialization: Add the starting point \((1, 1)\) to the queue with distance \(0\)

2. BFS Main Loop:

   • Dequeue the current position \((r, c)\)
   • For each of the 4 directions (up, down, left, right), calculate the position after sliding
     • Move forward until hitting a wall (edge of the rink) or stopping just before a pillar
   • If the destination is unvisited, record the distance and add it to the queue
   • If the goal is reached, output the distance and terminate

3. Sliding Process Details:

   From current position (nr, nc), move one cell at a time in direction (dr, dc)
   while True:
      next position = (nr + dr, nc + dc)
      if next position is outside the wall or there is a pillar at next position:
          break  # Stop at current position
      nr, nc = next position  # Continue moving
   

4. Termination Condition: If the queue becomes empty without reaching the goal, output -1

Complexity

• Time Complexity: \(O(H \times W \times \max(H, W))\)

  • Each cell enters the queue at most once: \(O(H \times W)\) states
  • For each state, sliding in 4 directions moves at most \(O(\max(H, W))\) cells

• Space Complexity: \(O(H \times W)\)

  • Size of the dictionary (or array) to record distances
  • Size of the set to record pillar positions: \(O(N)\)

With constraints \(H, W \leq 100\), the worst case computation is about \(10^6\), which runs sufficiently fast.

Implementation Tips

1. Manage pillar positions with a Set: Use a set to check whether there is a pillar in \(O(1)\) time.

2. Skip cases where sliding result is the same as original position: Avoid unnecessary processing when a wall is adjacent and you cannot actually move.

3. Early termination upon reaching the goal: In BFS, the first time you reach the goal is the shortest path, so you can output the answer immediately.

4. Coordinates are 1-indexed: By starting coordinates from 1 to match the problem statement, wall checks can be written intuitively as < 1 or > H (> W).

Source Code

from collections import deque

def solve():
    H, W, N = map(int, input().split())
    
    # Record pillar positions
    pillars = set()
    for _ in range(N):
        r, c = map(int, input().split())
        pillars.add((r, c))
    
    # Search for shortest path using BFS
    # State: (row, column)
    start = (1, 1)
    goal = (H, W)
    
    if start == goal:
        print(0)
        return
    
    # Record distances
    dist = {}
    dist[start] = 0
    
    queue = deque([start])
    
    # Directions: up, down, left, right
    directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
    
    while queue:
        r, c = queue.popleft()
        current_dist = dist[(r, c)]
        
        for dr, dc in directions:
            # Keep sliding in that direction
            nr, nc = r, c
            while True:
                next_r, next_c = nr + dr, nc + dc
                # Check if hitting a wall
                if next_r < 1 or next_r > H or next_c < 1 or next_c > W:
                    # Hit a wall, stop at current position
                    break
                # Check if there is a pillar
                if (next_r, next_c) in pillars:
                    # Stop just before the pillar
                    break
                # Can move
                nr, nc = next_r, next_c
            
            # Position after sliding
            if (nr, nc) != (r, c):  # If actually moved
                if (nr, nc) not in dist:
                    dist[(nr, nc)] = current_dist + 1
                    if (nr, nc) == goal:
                        print(current_dist + 1)
                        return
                    queue.append((nr, nc))
    
    # Cannot reach the goal
    print(-1)

solve()

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