原题链接在这里：https://leetcode.com/problems/minimum-cost-to-merge-stones/
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 = 2 Output: 20 Explanation: 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 = 3 Output: -1 Explanation: 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 = 3 Output: 25 Explanation: 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 <= 302 <= K <= 301 <= 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<j; m += K-1){
20 dp[i][j] = Math.min(dp[i][j], dp[i][m]+dp[m+1][j]);
21 }
22
23 if((size-1) % (K-1) == 0){
24 dp[i][j] += preSum[j+1] - preSum[i];
25 }
26 }
27 }
28
29 return dp[0][n-1];
30 }
31 }