【LeetCode 864 】Shortest Path to Get All Keys

Subject description:

Two-dimensional matrix, lowercase letters correspond to the key capital letters. There are walls and doors, not through the wall, only to have a key to the past when the door met. Q. least take many steps to get all the keys.

Ideas:

bfs, it began like an ordinary bfs the same record the number of steps each location. Later attempts to mark the number keys.
The original mark is to get all the keys. . . . Because only a maximum of six key can be used to determine whether the current bit operation have that key.

Code:

class Solution {
public:
    struct Node {
        int x, y;
        int status;
        Node(int x = 0, int y = 0, int status = 0) {
            this->x = x;
            this->y = y;
            this->status = status;
        }
    };
    
    int shortestPathAllKeys(vector<string>& grid) {
        int cnt = 0;
        int n = grid.size();
        int m = grid[0].length();
        int stx, sty;
        
        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                if (grid[i][j] >= 'a' && grid[i][j] <= 'z') {
                    cnt = max(cnt, grid[i][j] - 'a' + 1);
                }
                if (grid[i][j] == '@') {
                    stx = i, sty = j;
                }
            }
        }
        
        queue<Node> que;
        que.push(Node(stx, sty, 0));
        vector<vector<vector<int>> > step(n, vector<vector<int>>(m, vector<int>(1<<6, INT_MAX)));
        step[stx][sty][0] = 0;
        map<char, int> mp;

        int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        
        while(!que.empty()) {
            Node cur = que.front();
            int curx = cur.x;
            int cury = cur.y;
            int curstu = cur.status;
            que.pop();
                                                                                                        
            if (curstu == (1<<cnt)-1 ) return step[curx][cury][curstu];
                
            for (int i=0; i<4; ++i) {
                int nxtx = curx + dir[i][0];
                int nxty = cury + dir[i][1];
                if (nxtx < 0 || nxtx >= n || nxty < 0 || nxty >= m) continue;
                char nxt_ch = grid[nxtx][nxty];
                if (nxt_ch == '#') continue;
                if (nxt_ch >= 'a' && nxt_ch <= 'z') {
                    int tkey = nxt_ch-'a';
                    int nxtstu = curstu | (1<<tkey);
                
                    if (step[nxtx][nxty][nxtstu] > step[curx][cury][curstu] + 1) {
                        que.push(Node(nxtx, nxty, nxtstu));
                        
                        step[nxtx][nxty][nxtstu] = step[curx][cury][curstu] + 1;
                    }
                }
                if ((nxt_ch == '.' || nxt_ch =='@') && step[nxtx][nxty][curstu] > step[curx][cury][curstu] + 1) {
                    que.push(Node(nxtx, nxty, curstu));
                    step[nxtx][nxty][curstu] = step[curx][cury][curstu] + 1;
                }
                if (nxt_ch >= 'A' && nxt_ch <= 'Z') {
                    int key = nxt_ch - 'A';
                    if ((curstu & (1<<key)) && step[nxtx][nxty][curstu] > step[curx][cury][curstu] + 1) {
                        que.push(Node(nxtx, nxty, curstu));
                        step[nxtx][nxty][curstu] = step[curx][cury][curstu] + 1;
                    }
                }
            }
        }
        return -1;
    }
};

Morning: I'd like millet bracelet.
Experienced 1s sold out.