CF 102697106 - Geiger Counter

We have an n × m maze. Each cell contains a single digit from 0 to 9, representing the radiation level of that cell. We start at the upper-left cell and need to reach the bottom row. Movement is allowed between cells sharing a side, so we may move up, down, left, or right.

codeforcescompetitive-programming

CF 102697106 - Geiger Counter

Rating: -
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Solve time: 1m 40s
Verified: yes

Solution

Problem Understanding

We have an n × m maze. Each cell contains a single digit from 0 to 9, representing the radiation level of that cell. We start at the upper-left cell and need to reach the bottom row. Movement is allowed between cells sharing a side, so we may move up, down, left, or right.

The cost of a path is not the sum of the radiation levels. Instead, it is the largest value encountered anywhere along that path. We need the path whose largest cell value is as small as possible, and we output that minimum possible maximum.

The constraints make the graph interpretation useful. Both n and m are below 100, so there are fewer than 10,000 cells. A traversal of the entire maze is easily fast enough within the one-second limit. What is not feasible is examining an exponential number of possible paths. The maze has cycles, and even after observing that revisiting cells is unnecessary, the number of simple paths can still grow exponentially with the number of cells.

There are several boundary cases where a careless implementation can fail. If the maze consists of one cell, such as

1 1
7

the only possible path starts and ends there, so the answer is 7. An implementation that initializes its answer to 0 and only examines neighboring cells could incorrectly return 0.

The starting cell itself also contributes to the path's radiation. For example,

1 2
82

has only one possible path, so the answer is 8. An implementation that checks only cells entered after the start could incorrectly return 2.

A path does not need to be the geometrically shortest path. Consider

3 3
159
189
111

The direct-looking route through the upper row reaches 9, but the route around the bottom uses only values at most 1 after leaving the starting cell. Since the starting cell is 1, the correct answer is 1. An algorithm that minimizes the number of moves first and then evaluates that path can return the wrong value.

Equal values are another useful sanity check. For

2 3
555
555

every possible path has maximum radiation 5, so the answer is 5. The algorithm must not assume that the answer changes whenever it moves to another cell.

Approaches

A direct brute-force solution treats every possible route from the upper-left cell as a candidate. During a depth-first search, we can maintain the maximum radiation encountered so far and update the global answer whenever the bottom row is reached. Since revisiting a cell never helps, we can mark cells as visited during the current path and enumerate simple paths.

This is correct because every valid route is eventually generated, and each route is evaluated using exactly the maximum cell value along that route. The problem is the number of routes. If the maze has V = n*m cells, a simple upper bound comes from choosing at most four directions for the first move and at most three new directions at every later move, giving O(4 * 3^(V-1)) possible walks before the simple-path restriction is even considered. Thus the worst-case work is exponential in V. A DFS that processes each generated path once takes Θ(P) path-expansion work when the current maximum is maintained incrementally, where P is the number of simple paths. With nearly 10,000 cells, this is completely impractical.

The key observation is that the objective is a minimum possible maximum. Instead of asking directly for the best path, we can guess the maximum radiation we are willing to tolerate.

Suppose we choose a threshold x. Now forbid every cell whose radiation is greater than x. The question becomes much simpler: can we travel from the starting cell to the bottom row using only allowed cells?

That is an ordinary graph reachability problem. A BFS or DFS can answer it in O(nm) time. More importantly, the feasibility property is monotonic. If a threshold x allows a path, then every larger threshold also allows that path, because increasing the threshold only makes more cells available.

There are only ten possible thresholds because every cell contains a digit from 0 through 9. We can simply test thresholds in increasing order and stop at the first one that permits reaching the bottom row. This avoids even needing binary search and keeps the implementation very small.

The brute-force works because it explicitly considers every candidate route, but fails when the number of routes becomes exponential. The threshold observation lets us replace route optimization with repeated graph reachability, reducing the problem to at most ten traversals of a graph containing fewer than 10,000 vertices.

Approach Time Complexity Space Complexity Verdict
Brute Force Exponential, Θ(P) for P simple paths O(nm) Too slow
Threshold BFS O(10nm) = O(nm) O(nm) Accepted

Algorithm Walkthrough

  1. Read the maze and store every digit as an integer. We need integer comparisons against the candidate radiation threshold.
  2. Try thresholds from 0 through 9 in increasing order. A threshold t means that cells with radiation at most t may be entered, while cells greater than t are treated as blocked.
  3. Before running BFS for a threshold, check whether the starting cell has radiation greater than t. If it does, no path can possibly work because every path contains the starting cell.
  4. Run BFS from (0, 0). For each cell, inspect its four orthogonal neighbors and enqueue a neighbor exactly when it lies inside the maze, has radiation at most t, and has not already been visited.
  5. As soon as BFS reaches any cell in the last row, the current threshold is feasible. Because thresholds are tested from smallest to largest, this is the minimum possible maximum radiation, so output it immediately.
  6. If a threshold cannot reach the bottom row, discard it and continue with the next threshold. Since the largest cell value is at most 9, some threshold will always succeed, namely 9.

Why it works

For a fixed threshold t, BFS visits exactly the cells that belong to the connected component of the starting cell after all cells with radiation greater than t have been removed. Thus BFS reaches the bottom row if and only if there exists a path whose every cell has radiation at most t.

Let the optimal answer be k. Every threshold smaller than k is impossible, because such a threshold would imply the existence of a path with maximum radiation smaller than the optimum. Threshold k is possible by definition of the optimal path. Since the algorithm checks thresholds in increasing order and returns the first feasible one, it returns exactly k.

Python Solution

import sys
from collections import deque

input = sys.stdin.readline

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

    directions = ((1, 0), (-1, 0), (0, 1), (0, -1))

    for limit in range(10):
        if grid[0][0] > limit:
            continue

        visited = [[False] * m for _ in range(n)]
        visited[0][0] = True

        q = deque([(0, 0)])

        while q:
            r, c = q.popleft()

            if r == n - 1:
                print(limit)
                return

            for dr, dc in directions:
                nr = r + dr
                nc = c + dc

                if not (0 <= nr < n and 0 <= nc < m):
                    continue
                if visited[nr][nc]:
                    continue
                if grid[nr][nc] > limit:
                    continue

                visited[nr][nc] = True
                q.append((nr, nc))

if __name__ == "__main__":
    solve()

The grid is stored as integers so that grid[r][c] > limit directly expresses whether a cell is forbidden for the current BFS.

The outer loop considers all possible answers in sorted order. There is no need to search values outside 0 through 9, because those are the only radiation levels appearing in the maze.

For each threshold, visited records the cells already reached. A cell is marked when it is inserted into the queue rather than when it is removed. This prevents the same cell from being queued several times through different neighbors.

The check r == n - 1 is enough to detect success because reaching any cell in the last row means that the required destination has been reached. We do not need to continue searching after that point.

The boundary condition checks both row and column ranges before accessing the grid. This avoids negative-index behavior in Python, which is especially easy to overlook because an expression such as grid[-1][c] is valid Python but refers to the last row instead of being rejected.

There is no integer-overflow concern because all values are digits and Python integers have arbitrary precision anyway. The queue contains at most n*m cells for one BFS.

Worked Examples

For Sample 1, the maze is

6 8
13567657
24903578
83107213
98829363
25511282
39108443

The threshold search progresses until 4 becomes feasible.

Threshold Starting cell allowed Bottom row reached Result
0 No No Reject
1 Yes No Reject
2 Yes No Reject
3 Yes No Reject
4 Yes Yes Answer 4

At threshold 4, BFS can travel through cells whose values are at most 4 and eventually reaches the last row. Since every smaller threshold failed, 4 is optimal.

For Sample 2, the maze is

6 5
89888
88898
99998
88888
89999
88888

The starting cell has value 8, so thresholds below 8 can immediately be rejected.

Threshold Starting cell allowed Bottom row reached Result
0 No No Reject
1 No No Reject
2 No No Reject
3 No No Reject
4 No No Reject
5 No No Reject
6 No No Reject
7 No No Reject
8 Yes Yes Answer 8

This example demonstrates why the starting cell must be included in the path cost. The answer cannot be below 8, regardless of what happens elsewhere in the maze.

Complexity Analysis

Measure Complexity Explanation
Time O(nm) At most 10 BFS traversals are performed, each visiting every cell and checking four neighbors
Space O(nm) The visited matrix and BFS queue can each contain O(nm) cells

With n,m < 100, there are fewer than 10,000 cells. Even multiplying a full traversal by the ten possible radiation values gives only a small constant multiple of the number of cells, so the solution comfortably fits the one-second time limit and 256 MB memory limit.

Test Cases

import sys
import io
from collections import deque

def solve():
    input = sys.stdin.readline

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

    directions = ((1, 0), (-1, 0), (0, 1), (0, -1))

    for limit in range(10):
        if grid[0][0] > limit:
            continue

        visited = [[False] * m for _ in range(n)]
        visited[0][0] = True
        q = deque([(0, 0)])

        while q:
            r, c = q.popleft()

            if r == n - 1:
                print(limit)
                return

            for dr, dc in directions:
                nr = r + dr
                nc = c + dc

                if not (0 <= nr < n and 0 <= nc < m):
                    continue
                if visited[nr][nc]:
                    continue
                if grid[nr][nc] > limit:
                    continue

                visited[nr][nc] = True
                q.append((nr, nc))

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample 1
assert run("""6 8
13567657
24903578
83107213
98829363
25511282
39108443
""") == "4\n", "sample 1"

# Provided sample 2
assert run("""6 5
89888
88898
99998
88888
89999
88888
""") == "8\n", "sample 2"

# Minimum-size input
assert run("""1 1
7
""") == "7\n", "single cell"

# All cells equal
assert run("""3 4
5555
5555
5555
""") == "5\n", "all equal values"

# Starting cell is the limiting value
assert run("""1 2
82
""") == "8\n", "starting-cell boundary"

# A larger grid where the low-radiation route requires a detour
assert run("""3 3
159
189
111
""") == "1\n", "detour around high cells"

# Maximum-size input
n = 99
m = 99
large = f"{n} {m}\n" + "\n".join(["0" * m] * n) + "\n"
assert run(large) == "0\n", "maximum-size all-zero grid"
Test input Expected output What it validates
1 1 / 7 7 Minimum-size maze and inclusion of the starting cell
3 4 / all 5 5 All-equal values
1 2 / 82 8 Starting-cell boundary
3 3 / 159,189,111 1 Choosing a longer low-radiation route instead of a tempting direct route
99 × 99 all zeros 0 Maximum-size input and traversal performance

Edge Cases

The one-cell maze

1 1
7

is handled because threshold 7 allows the starting cell, and the first BFS operation sees that its row is already the bottom row. The algorithm prints 7. There is no special-case code for a one-cell maze, because the general reachability logic already covers it.

The starting-cell boundary

1 2
82

shows why the threshold check must include (0, 0). Thresholds 0 through 7 are rejected immediately because the starting cell has value 8. At threshold 8, both cells are allowed and BFS reaches the only row immediately, giving 8.

The detour case

3 3
159
189
111

shows the difference between minimizing path length and minimizing the maximum cell value. At threshold 1, the cells form a route from the start down and across the bottom row. BFS discovers that connected component and reaches the destination without touching either 5, 8, or 9, so the answer is 1.

The all-equal case

2 3
555
555

fails for every threshold below 5, because no cell can be entered at those thresholds. At threshold 5, every cell becomes reachable, so BFS reaches the bottom row and returns 5. The algorithm does not depend on there being distinct radiation values.

The maximum-size case uses a 99 × 99 maze filled with zeroes. Threshold 0 already permits every cell, so BFS reaches the bottom row during its first traversal and returns 0. Only 9,801 cells need to be processed, demonstrating why the repeated-threshold approach remains easily fast enough under the given bounds.

The central invariant in every one of these cases is the same: during the BFS for threshold t, every visited cell is reachable from the start through cells whose radiation is at most t, and every cell that can be reached under that restriction is eventually visited. That turns the original minimax path problem into a sequence of ordinary reachability checks.