864. Shortest Path to Get All Keys

We are given a 2-dimensional grid. "." is an empty cell, "#" is a wall, "@" is the starting point, ("a", "b", ...) are keys, and ("A", "B", ...) are locks.

We start at the starting point, and one move consists of walking one space in one of the 4 cardinal directions.  We cannot walk outside the grid, or walk into a wall.  If we walk over a key, we pick it up.  We can't walk over a lock unless we have the corresponding key.

For some 1 <= K <= 6, there is exactly one lowercase and one uppercase letter of the first K letters of the English alphabet in the grid.  This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet.

Return the lowest number of moves to acquire all keys.  If it's impossible, return -1.

Example 1:

Input: ["@.a.#","###.#","b.A.B"]
Output: 8

Example 2:

Input: ["@..aA","..B#.","....b"]
Output: 6

Note:

1. 1 <= grid.length <= 30
2. 1 <= grid[0].length <= 30
3. grid[i][j] contains only '.', '#', '@', 'a'-'f' and 'A'-'F'
4. The number of keys is in [1, 6].  Each key has a different letter and opens exactly one lock.

Approach #1: C++. [BFS]

class Solution {
public:
    int shortestPathAllKeys(vector<string>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        queue<int> q;
        vector<vector<vector<int>>> seen(m, vector<vector<int>>(n, vector<int>(64, 0)));
        int allKeys = 0;
        
        //Init
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                const char c = grid[i][j];
                if (c == '@') {
                    q.push((i << 16) | (j << 8));
                    seen[i][j][0] = 1;
                } else if (c >= 'a' && c <= 'f') {
                    allKeys |= (1 << (c - 'a')); 
                }
            }
        }
        
        int steps = 0;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        
        while (!q.empty()) {
            int size = q.size();
            while (size--) {
                int cur = q.front(); q.pop();
                int x = cur >> 16;
                int y = (cur >> 8) & 0xFF;
                int keys = cur & 0xFF;
                
                if (keys == allKeys) return steps;
                
                for (int i = 0; i < 4; ++i) {
                    int xx = x + dirs[i];
                    int yy = y + dirs[i+1];
                    int curKeys = keys;
                    if (xx < 0 || xx >= m || yy < 0 || yy >= n) continue;
                    const char c = grid[xx][yy];
                    if (c == '#') continue;
                    if (c >= 'A' && c <= 'F' && !(keys & (1 << (c - 'A')))) continue;
                    if (c >= 'a' && c <= 'f') curKeys |= 1 << (c - 'a');
                    if (seen[xx][yy][curKeys]) continue;
                    seen[xx][yy][curKeys] = 1;
                    q.push((xx << 16) | (yy << 8) | curKeys);
                }
            }
            steps++;
        }
        
        return -1;
    }
};

　　

Analysis:

seen[x][y][keys] : To store the position and the number of keys. If this state don't be traveled we can do next step, otherwise we skip this state.

allKeys : To represent the keys which we will collect in this problem. In this problem we use six binary numbers to represent all the keys at difference bit. 

such as : a -> 1 so it will be represented by 000001 and f -> f - 'a' = 6 so it will be represent by 100000. If we have the keys of a and f so we can use 100001 to represent that.

queue<int> q : To simulation the BFS.