Stone Game

Posted 2020-07-28 Sheryl Wang

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.

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

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

这是一道比较典型的区间类DP题目.如果从小到大合并,则这题很难选择.一个比较好的方式是自顶向下,先考虑最后一次合并的费用.每一次合并的附加费用是合并的两堆石块的总数目,即新石堆的石块数目.

我们可以枚举合并的位置,看哪个位置的成本最小,这个过程可以一直进行下去,直到一个石块的时候,合并成本是0.这中间因为枚举分割位,有非常多的重复区间.所以需要判断是否已经处理过. f的DP矩阵一开始初始化为-1,如果某个位置不是-1,则说明已经处理过，不必另外维护一个visited矩阵，另外，每次我们都要求区间和，也是一个重复性非常高的过程，可以提前求出．一个是直接求各个区间点的值，是二维矩阵，另外一个是先求前缀和数组，利用这个数组求区间和．显然后者空间复杂度低很多．代码如下:

class Solution:
    # @param {int[]} A an integer array
    # @return {int} an integer
    def stoneGame(self, A):
        # memory search 
        if not A:
            return 0
        f = [[-1]*len(A) for i in xrange(len(A))]
        for i in xrange(len(A)):
            f[i][i] = 0
        cost = [0]*(len(A)+1)
        for i in xrange(1, len(A)+1):
            cost[i] = cost[i-1] + A[i-1]
        return self.search(A, f, cost, 0, len(A) - 1 )
        
    def search(self, A, f, cost, start, end):
        if f[start][end] >= 0:
            return f[start][end]
        import sys    
        f[start][end] = sys.maxint
        for k in xrange(start,end):
           left = self.search(A, f, cost, start, k)
           right = self.search(A, f, cost,  k+1, end)
           f[start][end] = min(f[start][end], left + right + cost[end+1] - cost[start])
        
        return f[start][end]

时间复杂度为Ｏ(n^3),枚举两边和切分位置．空间复杂度为Ｏ(n^2),DP矩阵