<p>You are given two groups of points where the first group has <code>size<sub>1</sub></code> points, the second group has <code>size<sub>2</sub></code> points, and <code>size<sub>1</sub> &gt;= size<sub>2</sub></code>.</p>

<p>The <code>cost</code> of the connection between any two points are given in an <code>size<sub>1</sub> x size<sub>2</sub></code> matrix where <code>cost[i][j]</code> is the cost of connecting point <code>i</code> of the first group and point <code>j</code> of the second group. The groups are connected if <strong>each point in both groups is connected to one or more points in the opposite group</strong>. In other words, each point in the first group must be connected to at least one point in the second group, and each point in the second group must be connected to at least one point in the first group.</p>

<p>Return <em>the minimum cost it takes to connect the two groups</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/03/ex1.jpg" style="width: 322px; height: 243px;" />
<pre>
<strong>Input:</strong> cost = [[15, 96], [36, 2]]
<strong>Output:</strong> 17
<strong>Explanation</strong>: The optimal way of connecting the groups is:
1--A
2--B
This results in a total cost of 17.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/03/ex2.jpg" style="width: 322px; height: 403px;" />
<pre>
<strong>Input:</strong> cost = [[1, 3, 5], [4, 1, 1], [1, 5, 3]]
<strong>Output:</strong> 4
<strong>Explanation</strong>: The optimal way of connecting the groups is:
1--A
2--B
2--C
3--A
This results in a total cost of 4.
Note that there are multiple points connected to point 2 in the first group and point A in the second group. This does not matter as there is no limit to the number of points that can be connected. We only care about the minimum total cost.
</pre>

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

<pre>
<strong>Input:</strong> cost = [[2, 5, 1], [3, 4, 7], [8, 1, 2], [6, 2, 4], [3, 8, 8]]
<strong>Output:</strong> 10
</pre>

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

<ul>
	<li><code>size<sub>1</sub> == cost.length</code></li>
	<li><code>size<sub>2</sub> == cost[i].length</code></li>
	<li><code>1 &lt;= size<sub>1</sub>, size<sub>2</sub> &lt;= 12</code></li>
	<li><code>size<sub>1</sub> &gt;= size<sub>2</sub></code></li>
	<li><code>0 &lt;= cost[i][j] &lt;= 100</code></li>
</ul>

class Solution {
public:
    int connectTwoGroups(vector<vector<int>>& cost) {
        vector<vector<int>> f(13, vector<int>((1<<13), 1e8));
        const int n = cost.size(), m = cost.back().size();

        f[0][0] = 0;
        for(int i = 1; i <= n; i++){
            for(int j = 0; j < m; j++){
                //宜家連接i同j
                for(int bit_mask = 0; bit_mask < (1<<m); bit_mask++){
                    //如果當前呢個bitmask無選擇到 j
                    if(!(bit_mask&(1<<j)))continue;

                    //假設 f[i][bit_mask] 已經包括第i個點同第j 個點
                    //佢只會由以下三種state過來
                    //我地選最小果個

                    
                    // set b中第j個點 無任何edge了
                    f[i][bit_mask] = min(f[i][bit_mask], f[i][bit_mask^(1<<j)]+ cost[i - 1][j]);
                    // set a第i個point 無任何edge 同 set b 第j 個point 沒有任何edge 
                    f[i][bit_mask] = min(f[i][bit_mask], f[i - 1][bit_mask^(1<<j)]+ cost[i - 1][j]);
                    // set a中第i個點沒有任何edge了
                    f[i][bit_mask] = min(f[i][bit_mask], f[i - 1][bit_mask]+ cost[i - 1][j]);
                }
            }
        }
        return f[n][(1<<m) - 1];
    }
};
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