https://www.nowcoder.com/practice/54ab9865ce7a45968b126d6968a77f34?tpId=196&tqId=40529&ru=/exam/oj

从每个腐烂的苹果开始使用广度优先遍历（bfs）

class Solution {
    int n, m;
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};
    vector<vector<bool>> vis;
  public:
    int rotApple(vector<vector<int> >& grid) {
        // write code here
        m = grid.size(), n = grid[0].size();
        vis.assign(m, vector<bool>(n, false));

        // 将腐烂苹果放进queue
        queue<pair<int, int>> q;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 2) {
                    q.push({i, j});
                }
            }
        }

        int ret = 0;
        // 一层一层的遍历，广度优先遍历
        while (q.size()) {
            int sz = q.size();
            ret++;
            while (sz--) {
                auto [a, b] = q.front();
                q.pop();
                for (int k = 0; k < 4; ++k) {
                    int x = a + dx[k];
                    int y = b + dy[k];
                    if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] == 1) {
                        vis[x][y] = true;
                        q.push({x, y});
                    }
                }
            }
        }

        // 判断是否有永不腐烂的苹果
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1 && !vis[i][j]) {
                    return -1;
                }
            }
        }
        
        return ret - 1;
    }
};