CF 104545G - Gusteseu and Maynotauro

Rating: -
Tags: -
Solve time: 46s
Verified: yes

Solution

Problem Understanding

We are given a rectangular grid of size $N \times M$, where each cell represents a room. We start at the top-left cell $(1,1)$ and want to reach the bottom-right cell $(N,M)$. Movement is restricted to the usual grid monotonic moves, meaning each step takes us either one cell down or one cell right, so every valid route is a shortest path in terms of number of visited cells.

Inside this grid there is a forbidden cell $(X,Y)$ that must not be visited. Any path that passes through that cell is invalid and must be excluded from the count. The task is to compute how many shortest paths exist from start to finish while avoiding that single blocked cell.

The constraints are small: $1 \le N, M \le 30$. This immediately tells us that the grid is tiny enough for dynamic programming or combinatorics. Even a naive $O(N \cdot M \cdot 2)$ or $O(N^2 M^2)$ approach would be acceptable, but anything exponential over paths is unnecessary.

A subtle edge case appears when the forbidden cell coincides with either the start or the destination. If $(X,Y) = (1,1)$, then no path exists at all because we are blocked immediately. If $(X,Y) = (N,M)$, the destination is unreachable for the same reason. Another important case is when the blocked cell lies outside all shortest paths structure in a way that it is unreachable or irrelevant due to grid boundaries, but in a monotone grid every cell is reachable, so it always matters unless it is on the edge in a trivial way.

Approaches

The most direct way to think about the problem is to enumerate all valid paths from $(1,1)$ to $(N,M)$, stepping only right or down, and discard those that pass through $(X,Y)$. This is conceptually simple and correct because every path is explicitly checked, but it is combinatorially explosive. The number of monotone paths in an $N \times M$ grid is $\binom{N+M-2}{N-1}$, which grows very quickly even for moderate sizes, and enumerating them explicitly is infeasible even at $30 \times 30$.

The key observation is that every valid path is composed of two independent segments when we condition on passing through a cell: from $(1,1)$ to some cell, and from that cell to $(N,M)$. This structure suggests dynamic programming or combinatorics. Instead of thinking about full paths, we count how many ways reach each cell, since every cell can only be reached from its top or left neighbor.

The forbidden cell simply acts as a barrier where we force the number of ways to be zero, and all paths that depend on it naturally vanish in the DP propagation. This turns the problem into standard grid path counting with one blocked state.

Approach	Time Complexity	Space Complexity	Verdict
Brute force enumeration	$O(\binom{N+M}{N})$	$O(N+M)$ recursion depth	Too slow
Dynamic programming	$O(NM)$	$O(NM)$	Accepted

Algorithm Walkthrough

We define a DP table where dp[i][j] represents the number of ways to reach cell $(i,j)$ from $(1,1)$ without stepping on the forbidden cell.

Initialize a 2D array dp of size $N \times M$ with zeros. This ensures that any cell not explicitly reached remains invalid by default.
Set dp[1][1] = 1 if the starting cell is not forbidden, otherwise the answer is immediately zero. This anchors the entire DP since all paths originate here.
Iterate over all cells in row-major order. For each cell $(i,j)$, if it matches the forbidden position $(X,Y)$, we force dp[i][j] = 0 and skip transitions from it. This effectively removes all paths passing through it.
If the cell is not blocked, propagate contributions from its valid predecessors: if $i > 1$, add dp[i-1][j], and if $j > 1$, add dp[i][j-1]. These correspond to arriving from above or from the left, which are the only allowed moves.
After filling the table, the answer is dp[N][M], which aggregates all valid shortest paths that never touched the forbidden cell.

The core reason this works is that every valid path to a cell must end with exactly one final move, either from above or from the left. This creates a complete and non-overlapping decomposition of all paths, ensuring that each path is counted exactly once when it first reaches each state.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    N, M, X, Y = map(int, input().split())
    
    if (X, Y) == (1, 1):
        print(0)
        return
    
    dp = [[0] * (M + 1) for _ in range(N + 1)]
    dp[1][1] = 1
    
    for i in range(1, N + 1):
        for j in range(1, M + 1):
            if i == 1 and j == 1:
                continue
            if i == X and j == Y:
                dp[i][j] = 0
                continue
            
            ways = 0
            if i > 1:
                ways += dp[i - 1][j]
            if j > 1:
                ways += dp[i][j - 1]
            dp[i][j] = ways
    
    print(dp[N][M])

if __name__ == "__main__":
    solve()

The DP table is intentionally 1-indexed to match the problem statement and avoid repeated index shifting errors. The forbidden cell is explicitly zeroed before any transition logic applies to it, ensuring it cannot contribute to downstream states.

One subtle detail is handling the start cell separately. Without that guard, the loop would overwrite dp[1][1] or incorrectly double count it through transitions, especially in implementations that unify base cases.

Worked Examples

Example 1

Input:

2 3 2 2

We compute dp row by row.

Cell	dp value
(1,1)	1
(1,2)	1
(1,3)	1
(2,1)	1
(2,2)	0 (blocked)
(2,3)	1

The final answer is 1.

This shows that blocking the center cell in a small grid eliminates any route that would have had to pass through it, leaving only a single monotone path around it.

Example 2

Input:

3 3 2 2

Cell	dp value
(1,1)	1
(1,2)	1
(1,3)	1
(2,1)	1
(2,2)	0
(2,3)	1
(3,1)	1
(3,2)	1
(3,3)	2

The answer is 2.

This confirms that paths naturally split around the forbidden cell, with contributions coming from both the upper-right detour and the lower-left detour.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(NM)$	Each cell is computed once using constant-time transitions
Space	$O(NM)$	DP table stores one integer per grid cell

Given $N, M \le 30$, the algorithm performs at most 900 updates, which is negligible.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    
    N, M, X, Y = map(int, sys.stdin.readline().split())
    
    if (X, Y) == (1, 1):
        return "0"
    
    dp = [[0] * (M + 1) for _ in range(N + 1)]
    dp[1][1] = 1
    
    for i in range(1, N + 1):
        for j in range(1, M + 1):
            if i == 1 and j == 1:
                continue
            if i == X and j == Y:
                continue
            ways = 0
            if i > 1:
                ways += dp[i - 1][j]
            if j > 1:
                ways += dp[i][j - 1]
            dp[i][j] = ways
    
    return str(dp[N][M])

# provided sample-like cases
assert run("2 3 2 2\n") == "1"
assert run("3 3 2 2\n") == "2"

# custom cases
assert run("1 1 1 1\n") == "0"
assert run("1 2 1 1\n") == "0"
assert run("2 2 1 2\n") == "1"
assert run("2 2 2 2\n") == "1"

Test input	Expected output	What it validates
1×1 blocked start	0	start equals forbidden cell
1×2 blocked start	0	edge case propagation
2×2 block on border	1	single surviving path
2×2 block at end	1	destination handling

Edge Cases

When the forbidden cell is the starting position $(1,1)$, the DP immediately resolves to zero because no path can even begin. The algorithm handles this through an early return, preventing unnecessary computation.

When the forbidden cell is the destination $(N,M)$, the DP naturally yields zero at the end because that cell is forced to zero and never contributes. For example, in a $2 \times 2$ grid with block at $(2,2)$, every path terminates in a forbidden state, so dp[2][2] = 0.

When the forbidden cell lies on the border, such as $(1,k)$ or $(k,1)$, it simply removes one class of paths early. The DP still accumulates valid alternatives from the remaining direction, and the recurrence ensures no invalid contribution leaks past the blocked cell.