<p>You are given several <code>boxes</code> with different colors represented by different positive numbers.</p>

<p>You may experience several rounds to remove boxes until there is no box left. Each time you can choose some continuous boxes with the same color (i.e., composed of <code>k</code> boxes, <code>k &gt;= 1</code>), remove them and get <code>k * k</code> points.</p>

<p>Return <em>the maximum points you can get</em>.</p>

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

<pre>
<strong>Input:</strong> boxes = [1,3,2,2,2,3,4,3,1]
<strong>Output:</strong> 23
<strong>Explanation:</strong>
[1, 3, 2, 2, 2, 3, 4, 3, 1] 
----&gt; [1, 3, 3, 4, 3, 1] (3*3=9 points) 
----&gt; [1, 3, 3, 3, 1] (1*1=1 points) 
----&gt; [1, 1] (3*3=9 points) 
----&gt; [] (2*2=4 points)
</pre>

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

<pre>
<strong>Input:</strong> boxes = [1,1,1]
<strong>Output:</strong> 9
</pre>

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

<pre>
<strong>Input:</strong> boxes = [1]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= boxes.length &lt;= 100</code></li>
	<li><code>1 &lt;= boxes[i]&nbsp;&lt;= 100</code></li>
</ul>

class Solution {
public:
    int removeBoxes(vector<int>& boxes) {
        int dp[110][110][200];
        memset(dp, 0, sizeof dp);

        int n = boxes.size();
        for (int start = n-1; start >= 0; start--) {
            for (int end = 0; end <= n; end++) {
                if (start >= end) continue;
                int cnt = 0;
                while (start+cnt < end and boxes[start] == boxes[start+cnt]) cnt++;
                for (int k = 0; k <= n-cnt; k++) {
                    int split1 = start+cnt;
                    dp[start][end][k] = dp[start+cnt][end][0] + (k+cnt)*(k+cnt);
                    for (int split2 = split1; split2 < end; split2++) {
                        if (boxes[split2] == boxes[start]) {
                            dp[start][end][k] = max(dp[start][end][k], 
                            dp[split1][split2][0] + dp[split2][end][k+cnt]);
                        }
                    }
                }
            }
        }
        return dp[0][n][0];   
    }
};
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