原题链接在这里：

题目：

There are N piles of stones arranged in a row.  The i-th pile has stones[i] stones.

A move consists of merging exactly K consecutive piles into one pile, and the cost of this move is equal to the total number of stones in these K piles.

Find the minimum cost to merge all piles of stones into one pile.  If it is impossible, return -1.

Example 1:

Input: stones = [3,2,4,1], K = 2Output: 20Explanation: We start with [3, 2, 4, 1].We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1].We merge [4, 1] for a cost of 5, and we are left with [5, 5].We merge [5, 5] for a cost of 10, and we are left with [10].The total cost was 20, and this is the minimum possible.

Example 2:

Input: stones = [3,2,4,1], K = 3Output: -1Explanation: After any merge operation, there are 2 piles left, and we can't merge anymore.  So the task is impossible.

Example 3:

Input: stones = [3,5,1,2,6], K = 3Output: 25Explanation: We start with [3, 5, 1, 2, 6].We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6].We merge [3, 8, 6] for a cost of 17, and we are left with [17].The total cost was 25, and this is the minimum possible.

Note:

• 1 <= stones.length <= 30
• 2 <= K <= 30
• 1 <= stones[i] <= 100

题解：

Each merge step, piles number decreased by K-1. Eventually there is only 1 pile. n - mergeTimes * (K-1) == 1. megeTimes = (n-1)/(K-1). If it is not divisable, then it could not merge into one pile, thus return -1.

Let dp[i][j] denotes minimum cost to merge [i, j] inclusively.

m = i, i+1, ... j-1. Let i to m be one pile, and m+1 to j to certain piles. dp[i][j] = min(dp[i][m] + dp[m+1][j]).

In order to make i to m as one pile, [i,m] inclusive length is multiple of K. m moves K-1 each step.

If [i, j] is multiple of K, then dp[i][j] could be merged into one pile. dp[i][j] += preSum[j+1] - preSum[i].

return dp[0][n-1], minimum cost to merge [0, n-1] inclusively.

Time Complexity: O(n^3/K).

Space: O(n^2).

AC Java:

1 class Solution { 2     public int mergeStones(int[] stones, int K) { 3         int n = stones.length; 4         if((n-1)%(K-1) != 0){ 5             return -1; 6         } 7          8         int [] preSum = new int[n+1]; 9         for(int i = 1; i<=n; i++){10             preSum[i] = preSum[i-1] + stones[i-1];11         }12         13         int [][] dp = new int[n][n];14         for(int size = 2; size<=n; size++){15             for(int i = 0; i<=n-size; i++){16                 int j = i+size-1;17                 dp[i][j] = Integer.MAX_VALUE;18                 19                 for(int m = i; m

类似.