<p>Alice and Bob continue their games with piles of stones. There are a number of piles <strong>arranged in a row</strong>, and each pile has a positive integer number of stones <code>piles[i]</code>. The objective of the game is to end with the most stones.</p>

<p>Alice and Bob take turns, with Alice starting first.</p>

<p>On each player&#39;s turn, that player can take <strong>all the stones</strong> in the <strong>first</strong> <code>X</code> remaining piles, where <code>1 &lt;= X &lt;= 2M</code>. Then, we set <code>M = max(M, X)</code>. Initially, M = 1.</p>

<p>The game continues until all the stones have been taken.</p>

<p>Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">piles = [2,7,9,4,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">10</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>If Alice takes one pile at the beginning, Bob takes two piles, then Alice takes 2 piles again. Alice can get <code>2 + 4 + 4 = 10</code> stones in total.</li>
	<li>If Alice takes two piles at the beginning, then Bob can take all three piles left. In this case, Alice get <code>2 + 7 = 9</code> stones in total.</li>
</ul>

<p>So we return 10 since it&#39;s larger.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">piles = [1,2,3,4,5,100]</span></p>

<p><strong>Output:</strong> <span class="example-io">104</span></p>
</div>

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

<ul>
	<li><code>1 &lt;= piles.length &lt;= 100</code></li>
	<li><code>1 &lt;= piles[i]&nbsp;&lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    unordered_map<int, unordered_map<int,int>> mpa, mpb; // M, remain bmp len, val
    vector<vector<int>> dp, dp2;
    int game(vector<int>& piles, int idx, int M, bool turn){ // true=alice, false=bob
        int remain_len = piles.size()-idx;

        if(turn && dp[M][remain_len]!=-1){
            return dp[M][remain_len];
        }
        else if(!turn && dp2[M][remain_len]!=-1){
            return dp2[M][remain_len];
        }

        if(turn && remain_len>0){
            int max_a = 0;
            int cur_val = 0;
            for(int i=idx; i<min(idx+2*M, (int)piles.size()); i++){
                cur_val += piles[i];
                int new_m = max(i-(idx-1), M);

                int temp = game(piles, i+1, new_m, false);
                max_a = max(max_a, cur_val+max(0, temp));

            }
            return dp[M][remain_len] = max_a;
        }
        else if(remain_len>0){
            int min_b = INT_MAX;
            int cur_val = 0;
            for(int i=idx; i<min(idx+2*M, (int)piles.size()); i++){
                cur_val += piles[i];
                int new_m = max(i-(idx-1), M);
                int temp = game(piles, i+1, new_m, true);
                min_b = min(min_b, max(0, temp));

            }
            return dp2[M][remain_len] = min_b==INT_MAX?0:min_b;
        }
        return -1; 

    }

    int stoneGameII(vector<int>& piles) {
        dp.resize(101, vector<int>(101,-1));
        dp2.resize(101, vector<int>(101,-1));
        return game(piles, 0, 1, true);
    }
};
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