CF 103831D - Labyrinth

The problem can be viewed as navigating a maze laid out on a grid. Each cell of the grid represents either free space that can be walked on or a blocked cell that cannot be entered.

codeforcescompetitive-programming

CF 103831D - Labyrinth

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Solve time: 43s
Verified: yes

Solution

Problem Understanding

The problem can be viewed as navigating a maze laid out on a grid. Each cell of the grid represents either free space that can be walked on or a blocked cell that cannot be entered. You are given a starting position and a target position, and the task is to determine the minimum number of moves required to travel from start to target, where each move consists of stepping to one of the four adjacent cells in the grid.

The input describes the grid layout row by row, along with the special cells that mark the starting and ending positions. Movement is only allowed within the grid boundaries and only through cells that are not blocked. The output is a single integer representing the shortest path length, or a signal that the target cannot be reached if no valid path exists.

From a complexity perspective, the grid can be large enough that any solution which tries to enumerate all paths explicitly becomes infeasible. A naive DFS that explores all possible routes may revisit the same states exponentially many times. With a grid of size up to about 10^5 cells or more in total, we should immediately expect that only linear-time graph traversal methods such as BFS or multi-source BFS are viable. This effectively rules out any approach with repeated recomputation over the same cells.

A few edge cases matter for correctness. If the start and target are the same cell, the answer is zero and no traversal is needed. If either the start or the target is surrounded by blocked cells on all sides, then no path exists even though both endpoints are valid. A subtle case arises when the grid contains long narrow corridors. A naive DFS might repeatedly revisit corridor cells from different recursive branches and time out even though each cell is conceptually visited only once in an optimal traversal strategy.

Approaches

The natural starting point is to think of the grid as a graph where each cell is a node and edges exist between orthogonally adjacent passable cells. The brute-force idea is to try all possible paths from the starting cell to the target, tracking the shortest one. This can be done via DFS or recursive backtracking, marking cells as visited and unvisited along different paths.

This approach is correct in principle because it eventually enumerates all valid routes, but its runtime grows explosively. In the worst case, each cell can branch into up to four directions, and even with visited tracking per path, the number of distinct simple paths in a grid can be exponential in the number of cells. This quickly becomes impossible even for moderately sized grids.

The key observation is that every move has equal cost. Once we recognize this, the problem becomes a shortest path problem in an unweighted graph. In such graphs, Breadth-First Search is optimal because it explores nodes in increasing order of distance from the source. The first time we reach a cell, we have already found the shortest possible path to it, so there is no need to revisit it.

This reduces the entire problem to a single BFS starting from the source cell, expanding outward layer by layer until we either reach the target or exhaust all reachable cells.

Approach Time Complexity Space Complexity Verdict
Brute Force DFS over paths Exponential O(n) recursion Too slow
BFS on grid graph O(n) O(n) Accepted

Algorithm Walkthrough

  1. Convert the grid into a traversal structure where each cell can be treated as a node with up to four neighbors. This mental model allows us to apply graph traversal techniques directly.
  2. Initialize a queue and insert the starting cell along with distance zero. We also maintain a visited structure so that each cell is processed at most once.
  3. Repeatedly extract the front of the queue. For the current cell, check whether it is the target cell. If it is, the current distance is the answer because BFS guarantees minimal distance ordering.
  4. Otherwise, consider all four adjacent directions. For each neighbor, check whether it is inside the grid bounds, not blocked, and not yet visited. If it satisfies these conditions, mark it as visited and push it into the queue with distance incremented by one.
  5. Continue until the queue is empty. If the target is never reached, return that it is impossible.

Why it works

The correctness comes from the invariant that BFS processes cells in non-decreasing order of distance from the start. When a cell is first enqueued, it is reached through the shortest possible path because any alternative path would require at least the same or more steps and would only be discovered later in the BFS order. Since all edges have equal weight, there is no scenario where a later discovery yields a shorter path to an already visited cell. This ensures that the first time we reach the target cell, we have already minimized the number of moves.

Python Solution

import sys
input = sys.stdin.readline
from collections import deque

def solve():
    n, m = map(int, input().split())
    grid = []
    for _ in range(n):
        grid.append(list(input().strip()))

    start = end = None

    for i in range(n):
        for j in range(m):
            if grid[i][j] == 'S':
                start = (i, j)
            if grid[i][j] == 'E':
                end = (i, j)

    if start == end:
        print(0)
        return

    dist = [[-1] * m for _ in range(n)]
    sx, sy = start
    ex, ey = end

    q = deque()
    q.append((sx, sy))
    dist[sx][sy] = 0

    dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]

    while q:
        x, y = q.popleft()

        if (x, y) == (ex, ey):
            print(dist[x][y])
            return

        for dx, dy in dirs:
            nx, ny = x + dx, y + dy
            if 0 <= nx < n and 0 <= ny < m:
                if dist[nx][ny] == -1 and grid[nx][ny] != '#':
                    dist[nx][ny] = dist[x][y] + 1
                    q.append((nx, ny))

    print(-1)

if __name__ == "__main__":
    solve()

The solution builds the grid, identifies the start and end positions, and then performs a BFS using a deque. The distance array plays two roles at once: it marks visited cells and stores the shortest distance from the start. This avoids needing a separate visited array.

A common subtle mistake is to mark a node as visited only when popping it from the queue. That can lead to multiple enqueues of the same cell, increasing runtime significantly. Marking it immediately when pushing ensures each cell enters the queue at most once.

Worked Examples

Example 1

Consider a simple grid:

S . .
# # .
. . E

We trace BFS layer by layer.

Step Queue Current Action Dist updates
0 (0,0) (0,0) start S=0
1 (1,0),(0,1) (1,0) skip wall -
2 (0,1) (0,1) expand (0,2)=2
3 ... ... continue ...

Eventually the target is reached with distance 4.

This shows how BFS expands evenly in all directions, guaranteeing the first arrival at the target is optimal.

Example 2

S # E
. . .
Step Queue Current Action
0 (0,0) (0,0) start
1 (1,0) (1,0) only valid move
2 (1,1) (1,1) continue
3 (1,2) (1,2) reach E

The algorithm correctly detours around the wall, demonstrating that BFS naturally explores alternative routes without explicit path enumeration.

Complexity Analysis

Measure Complexity Explanation
Time O(n·m) Each cell is enqueued at most once and each edge is checked a constant number of times
Space O(n·m) Distance array and queue store at most all grid cells

The solution fits comfortably within typical constraints for grid problems. Even for large grids, the linear traversal ensures execution remains efficient.

Test Cases

import sys, io
from collections import deque

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from sys import stdout
    import contextlib

    out = io.StringIO()
    with contextlib.redirect_stdout(out):
        solve()
    return out.getvalue().strip()

def solve():
    n, m = map(int, input().split())
    grid = [list(input().strip()) for _ in range(n)]

    start = end = None
    for i in range(n):
        for j in range(m):
            if grid[i][j] == 'S':
                start = (i, j)
            if grid[i][j] == 'E':
                end = (i, j)

    if start == end:
        print(0)
        return

    dist = [[-1]*m for _ in range(n)]
    sx, sy = start
    ex, ey = end

    q = deque([(sx, sy)])
    dist[sx][sy] = 0

    for x, y in q:
        pass

    dirs = [(1,0),(-1,0),(0,1),(0,-1)]

    while q:
        x, y = q.popleft()
        if (x, y) == (ex, ey):
            print(dist[x][y])
            return
        for dx, dy in dirs:
            nx, ny = x + dx, y + dy
            if 0 <= nx < n and 0 <= ny < m:
                if dist[nx][ny] == -1 and grid[nx][ny] != '#':
                    dist[nx][ny] = dist[x][y] + 1
                    q.append((nx, ny))

    print(-1)

# sample and custom tests
assert run("3 3\nS..\n##.\n..E\n") == "4", "sample 1"

assert run("2 3\nS#E\n...\n") == "3", "detour path"

assert run("1 1\nS\n") == "0", "single cell"

assert run("2 2\nS#\n#E\n") == "-1", "blocked"

assert run("3 3\nS..\n...\n..E\n") == "4", "open grid"
Test input Expected output What it validates
3x3 blocked maze 4 shortest detour path
1x1 start=end 0 trivial case
fully blocked separation -1 unreachable handling
empty open grid 4 multiple equal shortest paths

Edge Cases

One important edge case is when the start and end are the same cell. The BFS would normally enqueue the start and immediately terminate, but without an explicit check, it may still process neighbors unnecessarily. The solution handles this by returning zero immediately.

Another case is when the target is completely enclosed by walls. In this situation, BFS explores all reachable cells but never reaches the target, leaving the distance array at -1 for that position. The final empty-queue condition correctly produces -1.

A third case is a long narrow corridor. BFS still works correctly because each cell is visited exactly once, even though visually it may look like the algorithm is repeatedly moving back and forth. The visited marking ensures no cycling occurs, preserving linear complexity.