864. 获取所有钥匙的最短路径

864. 获取所有钥匙的最短路径

题解：

1. bfs
2. 总共不超过6把钥匙，通过位运算保存钥匙

class Solution {
    
    public int shortestPathAllKeys(String[] grid) {
        int[][][] dist = new int[31][31][64];
        int n = grid.length, m = grid[0].length(), S = 0;
        for (int i = 0; i < n; i++) {
            dist[i] = new int[31][];
            for (int j = 0; j < m; j++) {
                dist[i][j] = new int[64];
                for (int k = 0; k < 64; k ++) {
                    dist[i][j][k] = Integer.MAX_VALUE;
                }
            }
        }
        Queue<Node> q = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                // 起点
                if (grid[i].charAt(j) == '@') {
                    dist[i][j][0] = 0;
                    q.offer(new Node(i, j, 0));
                } else if (grid[i].charAt(j) >= 'a' && grid[i].charAt(j) <= 'z') {
                // 钥匙总数
                    S++;
                }
            }
        }
        int[] dx = new int[]{-1, 0, 1, 0};
        int[] dy = new int[]{0, 1, 0, -1};
        while (!q.isEmpty()) {
            Node t = q.poll();
            int d = dist[t.x][t.y][t.s];
            // 从这个点出发，向上下左右，四个方向走
            for (int i = 0; i < 4; i++) {
                int nx = t.x + dx[i], ny = t.y + dy[i], ns = t.s;
                // 判断是否合法
                if (nx < 0 || nx >= n || ny < 0 || ny >= m || grid[nx].charAt(ny) == '#') continue;
                char c = grid[nx].charAt(ny);
                if (c >= 'a' && c <= 'z') {
                    ns |= 1 << (c - 'a');
                    if (dist[nx][ny][ns] > d + 1) {
                        dist[nx][ny][ns] = d + 1;
                        // 钥匙凑够了，立马return
                        if (ns == (1 << S) - 1) return d + 1;
                        q.offer(new Node(nx, ny, ns));
                    }
                } else if (c >= 'A' && c <= 'Z') {
                    // 判断此时有没有这个锁的钥匙
                    int isUnLock = ns & (1 << c - 'A');
                    if (isUnLock > 0 && dist[nx][ny][ns] > d + 1) {
                        dist[nx][ny][ns] = d + 1;
                        q.offer(new Node(nx, ny, ns));
                    }
                } else {
                    if (dist[nx][ny][ns] > d + 1) {
                        dist[nx][ny][ns] = d + 1;
                        q.offer(new Node(nx, ny, ns));
                    }
                }
            }
        }
        return -1;
    }

    static class Node {
        int x, y, s;

        public Node(int x, int y, int s) {
            this.x = x;
            this.y = y;
            this.s = s;
        }
    }
}