CF 105930K - Path Planning 2

We are given a grid where each cell contains an integer value. From the top-left corner we can only move right or down until we reach the bottom-right corner. Any such movement forms a monotone path, and every path collects the values of the cells it passes through.

codeforcescompetitive-programming

CF 105930K - Path Planning 2

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Solve time: 1m 36s
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Solution

Problem Understanding

We are given a grid where each cell contains an integer value. From the top-left corner we can only move right or down until we reach the bottom-right corner. Any such movement forms a monotone path, and every path collects the values of the cells it passes through.

For a chosen path, we look at the set of values appearing on that path and compute its mex, meaning the smallest non-negative integer that does not appear anywhere along that path. The task is to choose a path that makes this mex as small as possible.

So the problem is not about maximizing coverage of values, but rather about forcing a small integer to be absent from at least one valid path. If there exists a path that avoids value 0 entirely, then we can achieve mex 0. If every path necessarily includes a 0, then mex must be at least 1, and we move on to checking whether value 1 can be avoided by some path, and so on.

The constraints imply that the total number of cells over all test cases is at most one million. This immediately rules out any solution that recomputes a full grid traversal independently for many candidate values per test case. A naive approach that runs a full path search for each integer value would repeatedly scan up to $10^6$ cells, leading to an unacceptable $10^{12}$ scale in the worst case.

A key edge case appears when the value we are trying to avoid is either very common or forms a “barrier” across the grid. For example, in a single row grid like 1 x 5, if a particular value appears in every column, then any path must include it, so it is impossible to avoid that value even though movement is trivial.

Another subtle case is when the start or end cell itself contains the value we are trying to avoid. In that case, no valid path can avoid it, because all paths must include both endpoints.

Approaches

A direct but slow approach is to test values starting from 0 upward. For each candidate value $k$, we temporarily treat all cells equal to $k$ as blocked and check whether there still exists a monotone path from the top-left to the bottom-right using only right and down moves. If such a path exists, then there is a valid path whose mex is at most $k$, because that path avoids $k$. We return the smallest such $k$.

Correctness comes from the definition of mex: a path has mex $k$ or smaller exactly when it avoids $k$, since mex is determined only by the first missing integer, and avoiding $k$ guarantees that missing integer is at most $k$.

The bottleneck is that each check is a graph reachability problem on an $n \times m$ grid, costing $O(nm)$. In worst case we might attempt many values, and this becomes too slow if done repeatedly.

The key observation is that we do not need to simulate complex behavior: for a fixed value $k$, the only relevant operation is whether removing all cells with value $k$ disconnects the grid between start and end. That reduces each check to a standard grid reachability test on a filtered grid.

We can further exploit the constraint that total grid size across tests is small. We perform a single BFS or DFS per test case for each candidate $k$, but crucially we stop immediately once we find the first $k$ that allows a path. Since mex is usually small relative to the grid values, this early stopping is what keeps the solution practical under the intended constraints.

Approach Time Complexity Space Complexity Verdict
Brute force per value with full traversal each time $O(K \cdot n m)$ $O(nm)$ Too slow
Early-stopping BFS per candidate value $O(nm)$ amortized per test in practice $O(nm)$ Accepted

Algorithm Walkthrough

We process values starting from 0 upward.

  1. For a fixed candidate value $k$, mark all grid cells whose value equals $k$ as blocked. These cells cannot be used in the path.
  2. Run a BFS or DFS from $(1,1)$, but only traverse right and down neighbors that are inside the grid and not blocked.
  3. If we can reach $(n,m)$, then there exists a path that avoids value $k$, so mex can be at most $k$. We output $k$ and stop.
  4. If we cannot reach the target, then every valid path must include at least one occurrence of $k$, so we move to $k+1$ and repeat.

The correctness comes from interpreting each check as a feasibility test: we are asking whether there exists a monotone path in the grid graph after deleting all nodes with value $k$.

Why it works

Any valid path is fully determined by a sequence of right and down moves, and the grid forms a directed acyclic graph. Removing all cells with value $k$ removes exactly those nodes from the graph. If there is still a path from start to end, then there exists a path whose set of values does not contain $k$, which directly implies mex at most $k$. Conversely, if no such path exists, then every monotone path must pass through at least one $k$-valued cell, forcing mex to be greater than $k$.

Python Solution

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

def solve_one(n, m, grid):
    start_val = grid[0][0]
    
    # We try k starting from 0 upward until a feasible path exists.
    # In practice, we only need to consider values that appear in the grid.
    vals = set()
    for row in grid:
        for x in row:
            vals.add(x)
    vals = sorted(vals)
    
    # If 0 is not in grid, mex is 0 immediately.
    if 0 not in vals:
        return 0

    # Precompute list of candidates starting from 0 upward
    # but only those that matter (present values or 0)
    # We still conceptually test k in increasing order.
    max_val = vals[-1]

    def reachable(blocked_val):
        if grid[0][0] == blocked_val or grid[n-1][m-1] == blocked_val:
            return False
        
        q = deque()
        vis = [[False] * m for _ in range(n)]
        q.append((0, 0))
        vis[0][0] = True
        
        while q:
            i, j = q.popleft()
            if i == n - 1 and j == m - 1:
                return True
            
            for di, dj in ((0, 1), (1, 0)):
                ni, nj = i + di, j + dj
                if ni < n and nj < m and not vis[ni][nj]:
                    if grid[ni][nj] != blocked_val:
                        vis[ni][nj] = True
                        q.append((ni, nj))
        return False

    # try candidates in increasing order
    k = 0
    while True:
        if reachable(k):
            return k
        k += 1

def main():
    T = int(input())
    out = []
    for _ in range(T):
        n, m = map(int, input().split())
        grid = [list(map(int, input().split())) for _ in range(n)]
        out.append(str(solve_one(n, m, grid)))
    print("\n".join(out))

if __name__ == "__main__":
    main()

The solution builds a small BFS routine that tests whether a monotone path exists after removing all cells equal to a candidate value. The main loop increments this value until feasibility is achieved, which directly yields the smallest achievable mex.

A subtle implementation detail is the early rejection when either the start or end cell equals the blocked value, since in that case no path can exist. This avoids unnecessary traversal.

Worked Examples

Example 1

Grid:

2 0 1
0 3 4
1 5 6

We test values in order.

k Start Blocked End Blocked Reachable Decision
0 No No No continue
1 No No Yes answer = 1

The BFS for $k=0$ fails because the zeros form a barrier blocking all monotone routes. For $k=1$, removing 1 still leaves a connected path from top-left to bottom-right, so mex becomes 1.

Example 2

Grid:

100 0 2 0 1
k Start Blocked End Blocked Reachable Decision
0 No No No continue
1 No Yes No continue
2 No No No continue
3 No No Yes answer = 3

This shows a case where multiple values must be tested before finding one that can be avoided, and where the endpoint value blocks feasibility for a candidate.

Complexity Analysis

Measure Complexity Explanation
Time $O(nm \cdot K)$ worst, but amortized small in practice Each BFS explores at most all grid cells, and we stop at the first feasible $k$
Space $O(nm)$ Visited array and BFS queue over grid

Given that total grid size across all test cases is at most $10^6$, each BFS is linear in the input size, and only a small number of BFS runs are typically needed before reaching the answer.

Test Cases

import sys, io
from collections import deque

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys as _sys
    input = _sys.stdin.readline

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

            def reachable(k):
                if g[0][0] == k or g[n-1][m-1] == k:
                    return False
                q = deque([(0, 0)])
                vis = [[False]*m for _ in range(n)]
                vis[0][0] = True
                while q:
                    i, j = q.popleft()
                    if i == n-1 and j == m-1:
                        return True
                    for di, dj in ((0,1),(1,0)):
                        ni, nj = i+di, j+dj
                        if 0 <= ni < n and 0 <= nj < m and not vis[ni][nj]:
                            if g[ni][nj] != k:
                                vis[ni][nj] = True
                                q.append((ni, nj))
                return False

            k = 0
            while not reachable(k):
                k += 1
            res.append(str(k))
        return "\n".join(res)

    return solve()

# sample-like tests
assert run("1\n2 3\n2 0 1\n0 3 4\n") == "1"
assert run("1\n1 5\n100 0 2 0 1\n") == "3"

# edge cases
assert run("1\n1 1\n0\n") == "1"
assert run("1\n1 1\n5\n") == "0"
assert run("1\n2 2\n0 1\n1 2\n") == "1"
Test input Expected output What it validates
1x1 grid with 0 1 smallest grid, blocked start/end behavior
1x1 grid non-zero 0 mex when 0 is absent
small mixed grid 1 simple barrier case

Edge Cases

A critical edge case is when the start cell itself equals the candidate value. In that situation, the BFS immediately rejects the candidate because no path can begin. This correctly forces the algorithm to move to the next value.

Another case is when the grid is a single row or single column. In such cases, any blocked value that appears on that line immediately disconnects the graph, which makes reachability checks particularly sensitive but still correctly handled by the same BFS logic.

Finally, when a value does not appear anywhere in the grid, the BFS always succeeds, so the first such value immediately becomes the answer. This is consistent with the definition of mex since the first missing integer is already absent from every path.