Farthest Reach Game

Analysis

We have an $N \times M$ grid with the following characters:

• S: Starting position (the player).
• E: An empty cell.
• B: A building cell.

We also have a move limit $K$.
Our task:

1. Find all “stable” cells (except those on the last row) that can be reached from S using $\le K$ moves.
2. Among those reachable cells, print the cell(s) that have the farthest Manhattan distance from S.
3. If multiple cells share the same maximum distance, print them in top-to-bottom, left-to-right order.

How the game works

1. Stability

   • A cell $(r, c)$ is “stable” if immediately below it is a building (B), i.e. grid[r+1][c] == 'B', or if $r$ is the last row, $r = N-1$.
   • However, the problem statement says: “The last row is stable by default, but you cannot use these cells as your destination.”
     • We are allowed to pass through or stand on the last row, but we cannot treat them as a valid final answer.

2. Gravity

   • If you ever land on (or move into) a cell that is not stable (and is not a B), you “slide/fall” straight downward until you reach a stable cell or go out of bounds.
   • Every single cell you traverse (including the starting cell and each cell as you fall) counts as 1 move.
   • If you fall out of the grid or onto a B, that path is invalid (you cannot remain on a B, nor can you go below row $N-1$).

3. Movement

   • You can move up/down/left/right one cell at a time, provided you do not end on a B.
   • Each step costs 1 move.
   • After each step, if you land on a non-stable empty cell, gravity kicks in, causing you to fall down until stable (each downward cell adds +1 to your move count).
   • You stop falling once you reach a stable E or the last row (E), or you discover it’s impossible (e.g. you hit a B directly below or go out of bounds).

4. Counting moves

   • Every single cell you “enter” in this process (horizontal/vertical/falling) increments the move count by 1.
   • You cannot exceed $K$ total moves.

5. Manhattan distance

   • The distance between $(r_1, c_1)$ and $(r_2, c_2)$ is $|r_1 - r_2| + |c_1 - c_2|$.

6. Output

   • Collect all stable cells (except in the last row) that you can reach with $\le K$ moves.
   • Find the maximum Manhattan distance (from S) among those cells.
   • Print all such max-distance cells, in order (row-major: top-to-bottom, then left-to-right).
   • If no stable cell (other than last row) is reachable, you would print nothing.

The trickiest part is faithfully implementing the movement + gravity rules in a way that matches the examples’ interpretation:

• From the examples, it appears each horizontal/vertical step costs 1,
• Falling through each cell downward also costs 1 per cell.
• You cannot stop on B, nor can you pass “through” it vertically. Hitting a B from above is effectively blocked.

Below is a step-by-step outline (followed by code):

Step-by-step Approach

1. Parse input

   • Read $N, M$.
   • Read the grid into a 2D array grid[N][M].
   • Find the start cell $(s_r, s_c)$ where grid[s_r][s_c] == 'S'.
   • Read $K$.

2. Identify stable cells

   • We mark a cell $(r,c)$ as “stable” if:
     1. $r = N-1$ (last row), or
     2. $r < N-1$ and grid[r+1][c] == 'B'.
   • However, recall that we cannot finalize on the last row $(N-1)$, but we can use it as intermediate.

3. Breadth-first search (BFS) with move-count tracking

   • We will maintain a 2D array dist[N][M], initialized to some large number (e.g. $\infty$), meaning “not visited yet or not reached yet.”
   • We will do a multi-step BFS where each push/pop from the queue tracks how many moves have been used so far.
   • Each “transition” can be up to 1 cell up/down/left/right, but we then apply gravity if needed.
   • For BFS we can use either:
     • A queue and carefully manage partial steps, or
     • A priority queue (Dijkstra-like) since each edge effectively has cost 1 (but we do a series of 1-cost steps, including falls).
   • However, given the constraints ($N, M \le 50$, $K \le 50$), a well-structured plain BFS can work if we carefully expand step-by-step.

4. How to expand a state

   • Let our state be $(r, c, used)$ = current cell and how many moves used so far.
   • From $(r, c)$, we attempt moving one cell up/down/left/right (call this the “intended next cell” $(nr, nc)$), provided:
     • It’s in-bounds,
     • It is not a B (we cannot “land” on building).
   • The cost to do that single step is +1. Let nextUsed = used + 1.
   • If nextUsed > K, we skip.
   • Now, if $(nr, nc)$ is a stable cell (or is last row), we can remain there for the moment.
   • If $(nr, nc)$ is not stable, we must fall:
     • Keep going down from $(nr, nc)$ while in-bounds, not landing on B, until either we find a stable E or fall out of bounds.
     • Each downward cell also costs +1 move.
     • If at any point the move cost exceeds $K$, we abandon that path.
   • Once we settle on a stable cell $(finalR, finalC)$ after gravity, we update dist[finalR][finalC] if nextUsedSoFar < dist[finalR][finalC]. Then push $(finalR, finalC)$ into the BFS queue with that cost.

5. Collect valid answers

   • After BFS completes, gather all stable cells $(r, c)$ with dist[r][c] \le K and $r \neq N-1$.
   • Compute Manhattan distance from $(s_r, s_c)$ to each such $(r,c)$.
   • Find the maximum distance. Then print all cells that attain that distance, in the required order.

6. Check Examples

   • The provided examples match this interpretation if we count each cell visited—including each step in the fall—as +1 to the move cost.

C++ Implementation

cpp
#include <bits/stdc++.h>
using namespace std;

// A large value for initialization
static const int INF = 1e9;

struct Cell {
    int r, c, cost;
};

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M;
    cin >> N >> M;

    vector<vector<char>> grid(N, vector<char>(M));
    int startR = -1, startC = -1;

    // Read the grid
    for(int r = 0; r < N; r++){
        for(int c = 0; c < M; c++){
            cin >> grid[r][c];
            if(grid[r][c] == 'S'){
                startR = r;
                startC = c;
            }
        }
    }

    int K;
    cin >> K;

    // Mark stable cells (including last row).
    // stable[r][c] = true if it's "E" and either (r == N-1) or grid[r+1][c] == 'B'.
    // We'll allow last row in BFS but won't count it as final "answer."
    vector<vector<bool>> stable(N, vector<bool>(M, false));
    for(int r = 0; r < N; r++){
        for(int c = 0; c < M; c++){
            if(grid[r][c] == 'B') {
                // Buildings are never stable "landing" spots.
                stable[r][c] = false;
            }
            else {
                if(r == N-1) {
                    // last row is stable by definition
                    stable[r][c] = true;
                }
                else if(grid[r+1][c] == 'B'){
                    stable[r][c] = true;
                }
                // else remains false
            }
        }
    }

    // dist[r][c] = minimum number of moves to land stably on (r,c)
    vector<vector<int>> dist(N, vector<int>(M, INF));
    dist[startR][startC] = 0;

    // We'll do a BFS (or D' BFS) that expands states in order of ascending cost.
    // But given K <= 50, a plain queue expansion step-by-step is also fine.
    // We'll here do a typical BFS with a queue of (r, c).
    queue<pair<int,int>>q;
    q.push({startR, startC});

    // Helper lambda: check in-bounds
    auto inBounds = [&](int rr, int cc){
        return (rr >= 0 && rr < N && cc >= 0 && cc < M);
    };

    while(!q.empty()){
        auto [r, c] = q.front();
        q.pop();
        int currCost = dist[r][c];
        
        // If we've already used up all K moves, no further expansion
        if(currCost > K) continue;

        // Try moving in 4 directions by 1 cell
        static int dr[4] = {-1, 1, 0, 0};
        static int dc[4] = {0, 0, -1, 1};

        for(int i=0; i<4; i++){
            int nr = r + dr[i];
            int nc = c + dc[i];
            int nextCost = currCost + 1;  // one step

            // Check bounds
            if(!inBounds(nr, nc)) continue;
            // We cannot land on a building
            if(grid[nr][nc] == 'B') continue;
            if(nextCost > K) continue;

            // Now if (nr,nc) is stable, we can just stand there
            // but if it's not stable, we must fall
            int finalR = nr;
            int finalC = nc;
            int fallCost = 0;

            // While this cell is not stable, keep falling
            while(finalR < N && !stable[finalR][finalC]) {
                // Attempt to move down by 1
                if(finalR+1 >= N) {
                    // out of bounds => invalid
                    finalR = -1;
                    break;
                }
                if(grid[finalR+1][finalC] == 'B') {
                    // building directly below => cannot fall onto building
                    // means we're stuck in a non-stable cell above a B, that violates the rules
                    finalR = -1;
                    break;
                }
                // otherwise we fall one step
                finalR++;
                fallCost++;
            }

            // If we fell out of bounds or onto a building, skip
            if(finalR == -1) {
                continue;
            }

            // total cost after falling
            int totalCost = nextCost + fallCost;
            if(totalCost > K) continue;

            // Now we're presumably on a stable cell or last row
            if(totalCost < dist[finalR][finalC]) {
                dist[finalR][finalC] = totalCost;
                q.push({finalR, finalC});
            }
        }
    }

    // Now gather all stable cells (except row N-1) that dist[r][c] <= K
    // and find the maximum manhattan distance from (startR, startC).
    int maxDist = -1;
    vector<pair<int,int>> candidates;

    for(int r=0; r<N-1; r++){  // exclude last row
        for(int c=0; c<M; c++){
            if(stable[r][c] && dist[r][c] != INF && dist[r][c] <= K){
                // compute Manhattan distance from (startR, startC)
                int manh = abs(r - startR) + abs(c - startC);
                if(manh > maxDist){
                    maxDist = manh;
                    candidates.clear();
                    candidates.push_back({r, c});
                }
                else if(manh == maxDist){
                    candidates.push_back({r, c});
                }
            }
        }
    }

    // If maxDist is still -1, it means no stable cell was reachable (other than last row).
    // The problem statement’s examples do not clarify if we print nothing in that case.
    // We'll follow the instructions: "Print the cell(s) that are farthest..."
    // So if none is reachable, we simply print nothing.
    if(maxDist < 0){
        // No output
        return 0;
    }

    // Otherwise print all (r, c) with that manhattan distance
    // in order top-to-bottom, left-to-right.
    // Our candidates are already in row-major order because we traversed r from top to bottom,
    // and c from left to right, so just print them.
    for(auto &cell : candidates){
        cout << cell.first << " " << cell.second << "\n";
    }

    return 0;
}

Explanation of Key Points

1. dist[r][c] tracks the minimum moves required to arrive stably on $(r,c)$.
2. For every expansion from $(r,c)$:
   • We try to move 1 step up/down/left/right (cost = +1).
   • If the new cell is not stable, we fall downward, incrementing the cost by +1 per row we fall.
   • We stop falling if/when we hit a stable cell or exit the board / land on a B.
   • If the final cost \le K, we update dist and push the resulting stable cell into the BFS queue.
3. At the end, we filter out:
   • Cells on the last row, since they cannot be final destinations (though we do use them for intermediate stepping).
   • Cells that are unreachable (dist == INF or dist > K).
4. Among the remaining stable cells, we pick the ones with the largest Manhattan distance from S and print them in ascending (row, col) order.

This matches the problem’s examples and constraints.