<p>You are given an <code>m x n</code> matrix <code>mat</code> that has its rows sorted in non-decreasing order and an integer <code>k</code>.</p>

<p>You are allowed to choose <strong>exactly one element</strong> from each row to form an array.</p>

<p>Return <em>the </em><code>k<sup>th</sup></code><em> smallest array sum among all possible arrays</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 5
<strong>Output:</strong> 7
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 9
<strong>Output:</strong> 17
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
<strong>Output:</strong> 9
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.  
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat.length[i]</code></li>
	<li><code>1 &lt;= m, n &lt;= 40</code></li>
	<li><code>1 &lt;= mat[i][j] &lt;= 5000</code></li>
	<li><code>1 &lt;= k &lt;= min(200, n<sup>m</sup>)</code></li>
	<li><code>mat[i]</code> is a non-decreasing array.</li>
</ul>

class Solution {
public:
    struct node{
        int s=0;
        vector<int> I;
        node() {
            s=0;
        }
    };
    int kthSmallest(vector<vector<int>>& mat, int k) {
        int m=mat.size(), n=mat[0].size(), i, j;
        auto cmp=[&mat](auto a, auto b) {
            return a.s>b.s;
        };
        priority_queue<node, vector<node>, decltype(cmp)> q(cmp);
        node s;
        for(i=0; i<m; i++) s.s+=mat[i][0];
        s.I.resize(m);
        q.push(s);
        set<vector<int>> v;
        v.insert(s.I);
        while(q.size() && k) {
            auto a=q.top();
            q.pop();
            if (k==1) return a.s;
            k--;
            for(i=0; i<m; i++) {
                if (a.I[i]==n-1) continue;
                auto na=a;
                na.I[i]++;
                if (v.count(na.I)) continue;
                na.s+=mat[i][na.I[i]]-mat[i][na.I[i]-1];
                q.push(na);
                v.insert(na.I);
            }
        }
        return -1;
    }
};
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