Amazing (Maze Exploration)

Problem Statement Summary

This is an interactive problem. You are placed in an $R \times C$ grid maze whose structure is unknown. At each step you can query which directions (N, S, E, W) are open from the current cell. Navigate from a start cell to a goal cell using as few moves as possible.

Solution: BFS-Based Exploration

Strategy

Since the maze is initially unknown, we must explore it on the fly. We maintain a map of all explored cells and their open walls. At each step:

1. If the current cell is unvisited, query its open directions and record them.

2. Run BFS on the known graph to plan a route:

   • If the goal is reachable through known cells, follow the shortest known path.

   • Otherwise, navigate toward the nearest unvisited cell reachable from the known graph.

   • Execute one step of the planned route. enumerate

     Correctness

     Every cell is visited at most once (we never re-query a visited cell). Each navigation step either visits a new cell or moves along a shortest path in the known graph. Since BFS always targets the nearest frontier cell, every reachable cell is eventually explored.

     Alternative: Tremaux's Algorithm

     For a DFS-based exploration, Tremaux's algorithm guarantees visiting each edge at most twice, giving $O(R \times C)$ total moves with simpler bookkeeping.

     C++ Implementation

     #include <bits/stdc++.h>
     using namespace std;
     
     const int dx[] = {-1, 1, 0, 0}; // N, S, E, W
     const int dy[] = {0, 0, 1, -1};
     const char dirChar[] = {'N', 'S', 'E', 'W'};
     
     int main() {
         int R, C, sr, sc, gr, gc;
         cin >> R >> C >> sr >> sc >> gr >> gc;
     
         int cx = sr, cy = sc;
     
         auto cellKey = [](int x, int y) -> long long {
             return ((long long)x << 32) | (unsigned)y;
         };
     
         unordered_map<long long, vector<int>> openDirs;
         unordered_set<long long> visited;
     
         while (cx != gr || cy != gc) {
             long long key = cellKey(cx, cy);
     
             if (!visited.count(key)) {
                 visited.insert(key);
                 string dirs;
                 cin >> dirs;
                 vector<int> open;
                 for (char c : dirs)
                     for (int d = 0; d < 4; d++)
                         if (c == dirChar[d])
                             open.push_back(d);
                 openDirs[key] = open;
             }
     
             // BFS to find goal or nearest unvisited cell
             queue<pair<int, int>> q;
             unordered_map<long long, long long> parent;
             unordered_map<long long, int> parentDir;
             q.push({cx, cy});
             parent[key] = -1;
     
             int tx = cx, ty = cy;
             bool found = false;
     
             while (!q.empty()) {
                 auto [x, y] = q.front(); q.pop();
                 long long ck = cellKey(x, y);
     
                 if (x == gr && y == gc) {
                     tx = x; ty = y; found = true; break;
                 }
     
                 if (!openDirs.count(ck)) continue;
                 for (int d : openDirs[ck]) {
                     int nx = x + dx[d], ny = y + dy[d];
                     long long nk = cellKey(nx, ny);
                     if (parent.count(nk)) continue;
                     parent[nk] = ck;
                     parentDir[nk] = d;
                     if (!visited.count(nk) && !found) {
                         tx = nx; ty = ny; found = true;
                     }
                     q.push({nx, ny});
                 }
             }
     
             // Reconstruct path from (cx,cy) to (tx,ty) and take first step
             vector<int> path;
             long long tk = cellKey(tx, ty);
             while (tk != cellKey(cx, cy)) {
                 path.push_back(parentDir[tk]);
                 tk = parent[tk];
             }
             reverse(path.begin(), path.end());
     
             if (!path.empty()) {
                 cout << dirChar[path[0]] << endl;
                 cx += dx[path[0]];
                 cy += dy[path[0]];
             }
         }
     
         return 0;
     }

     Complexity Analysis

     • Total moves: $O(R \times C)$---each cell is visited at most a constant number of times.

     • Time per step: $O(R \times C)$ for BFS over the known graph. Total computation: $O((R \times C)^2)$ worst case, but typically much better.

     • Space: $O(R \times C)$ for the explored maze graph.