There is a stone game. At the beginning of the game the player picks n piles of stones in a line.

The goal is to merge the stones in one pile observing the following rules:

  1. At each step of the game,the player can merge two adjacent piles to a new pile.
  2. The score is the number of stones in the new pile.

You are to determine the minimum of the total score.

Example
For [4, 1, 1, 4], in the best solution, the total score is 18:

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3
1. Merge second and third piles => [4, 2, 4], score +2
2. Merge the first two piles => [6, 4]，score +6
3. Merge the last two piles => [10], score +10

Other two examples:
[1, 1, 1, 1] return 8
[4, 4, 5, 9] return 43

自顶向下
动态规划四要素：

  1. State: dp[i][j] 表示把第i到第j个石子合并到一起的最小花费
  2. Function: dp[i][j] = min(dp[i][k] + dp[k + 1][j] + sum[i][j]), k 属于 [i, j)
  3. Initialize:
  4. Answer: dp[0][n - 1]
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public int stoneGame(int[] A) {
    if (A == null || A.length == 0)
        return 0;
    int n = A.length;
    int[][] dp = new int[n][n];
    boolean[][] visited = new boolean[n][n];
    int[][] sum = new int[n][n];
    for (int i = 0; i < n; i++) {
        sum[i][i] = A[i];
        for (int j = i + 1; j < n; j++) {
            sum[i][j] = sum[i][j - 1] + A[j];
        }
    }
    return search(0, n - 1, A, dp, visited, sum);
}
public int search(int l, int r, int[] A, int[][] dp, boolean[][] visited, int[][] sum) {
    if (l == r) {
        return 0;
    } else {
        if (visited[l][r])
            return dp[l][r];
        int min = Integer.MAX_VALUE;
        for (int i = l; i < r; i++) {
            min = Math.min(search(l, i, A, dp, visited, sum) + search(i + 1, r, A, dp, visited, sum) + sum[l][r], min);
        }
        dp[l][r] = min;
        visited[l][r] = true;
        return dp[l][r];
    }
}