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# 1000.Minimum-Cost-to-Merge-Stones

## 题目描述

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 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 .
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 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 .
The total cost was 25, and this is the minimum possible.
Note:
1 <= stones.length <= 30
2 <= K <= 30
1 <= stones[i] <= 100

## 代码

### Approach #1

class Solution {
public int mergeStones(int[] stones, int K) {
int n = stones.length;
if ((n - 1) % (K - 1) != 0) return -1;
int[] sum = new int[n + 1];
for (int i = 0; i < n; i++) {
sum[i + 1] = sum[i] + stones[i];
}
int[][] dp = new int[n][n];
for (int len = K; len <= n; len++) {
for (int i = 0; i + len <= n; i++) {
int j = i + len - 1;
dp[i][j] = Integer.MAX_VALUE;
for (int mid = i; mid < j; mid += K - 1) {
dp[i][j] = Math.min(dp[i][j], dp[i][mid] + dp[mid + 1][j]);
}
if ((j - i) % (K - 1) == 0) {
dp[i][j] += sum[j + 1] - sum[i];
}
}
}
return dp[n - 1];
}
}