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;
}
};