LintCode "Kth Smallest Number in Sorted Matrix"

1AC! Yes! Intuitive thought will be: we walk from up-left to down-right - but we should only walk along the 'lowest edge' in the sorted matrix - so it is min-heap based BFS solution. Tricky, and really fun!

class Solution {
    struct Item
    {
    Item(): val(0), x(0), y(0){};
    Item(int v, int xx, int yy): val(v), x(xx), y(yy){};
    int val;
    int x;
    int y;
    //
    bool operator()(const Item &r1, const Item &r2) const
    {
        return r1.val > r2.val;
    }
    };
public:
    /**
     * @param matrix: a matrix of integers
     * @param k: an integer
     * @return: the kth smallest number in the matrix
     */
    int kthSmallest(vector<vector<int> > &m, int k) {
    int h = m.size();
    int w = m[0].size();
    vector<vector<bool>> rec(h, vector<bool>(w, false));
    
        priority_queue<Item, std::vector<Item>, Item> q;
    int i = 0, j = 0;
    Item ret;
    
    q.push(Item(m[i][j], j, i));
    rec[i][j] = true;
    while(k)
    {
        ret = q.top();q.pop();
        j = ret.x; i = ret.y;
        if(i < (h - 1) && !rec[i + 1][j])
        {
        q.push(Item(m[i + 1][j], j, i + 1));
        rec[i + 1][j] = true;        
        }
        if(j < (w - 1) && !rec[i][j + 1])
        {
        q.push(Item(m[i][j + 1], j + 1, i));
        rec[i][j + 1] = true;        
        }
        k--;
    }
    return ret.val;
    }
};