Problem:
Score : 600 points

Problem Statement

We will create an artwork by painting black some squares in a white square grid with 10^9 rows and N columns.

The current plan is as follows: for the i-th column from the left, we will paint the H_i bottommost squares and will not paint the other squares in that column.

Before starting to work, you can choose at most K columns (possibly zero) and change the values of H_i for these columns to any integers of your choice between 0 and 10^9 (inclusive).

Different values can be chosen for different columns.

Then, you will create the modified artwork by repeating the following operation:

Choose one or more consecutive squares in one row and paint them black. (Squares already painted black can be painted again, but squares not to be painted according to the modified plan should not be painted.)

Find the minimum number of times you need to perform this operation.

Constraints

1 \leq N \leq 300

0 \leq K \leq N

0 \leq H_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
H_1 H_2 ... H_N

Output

Print the minimum number of operations required.

Sample Input 1

4 1
2 3 4 1

Sample Output 1

3

For example, by changing the value of H_3 to 2, you can create the modified artwork by the following three operations:

Paint black the 1-st through 4-th squares from the left in the 1-st row from the bottom.

Paint black the 1-st through 3-rd squares from the left in the 2-nd row from the bottom.

Paint black the 2-nd square from the left in the 3-rd row from the bottom.

Sample Input 2

6 2
8 6 9 1 2 1

Sample Output 2

7

Sample Input 3

10 0
1 1000000000 1 1000000000 1 1000000000 1 1000000000 1 1000000000

Sample Output 3

4999999996

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <inttypes.h>

int N, K;
int H[512];

/* [最初, 2番目) */
int64_t sakugenTable[512][512];

int64_t memo[512][512];

int64_t search(int pos, int stock) {
	int64_t answer;
	int i;
	if (pos > N) return 0;
	if (memo[pos][stock]) return ~memo[pos][stock];

	/* 使わない */
	answer = search(pos + 1, stock);
	/* 使う */
	for(i = 1; i <= stock && pos + i <= N + 1; i++) {
		int64_t candidate = search(pos + i + 1, stock - i) + sakugenTable[pos][pos + i];
		if (candidate > answer) answer = candidate;
	}

	return ~(memo[pos][stock] = ~answer);
}

int main(void) {
	int i, j;
	int64_t all_cost;
	if (scanf("%d%d", &N, &K) != 2) return 1;
	for (i = 1; i <= N; i++) {
		if (scanf("%d", &H[i]) != 1) return 1;
	}
	H[0] = H[N + 1] = 0;
	for (i = 1; i <= N; i++) {
		int64_t sum = 0;
		for (j = i; j <= N + 1; j++) {
			if (H[j] < H[j - 1]) sum += H[j - 1] - H[j];
			sakugenTable[i][j] = sum + H[j];
		}
	}
	all_cost = sakugenTable[1][N + 1];
	for (i = 1; i <= N; i++) {
		for (j = i; j <= N + 1; j++) {
			int64_t after_cost = (H[i - 1] >= H[j]) ? H[i - 1] : H[j];
			sakugenTable[i][j] -= after_cost;
		}
	}
	printf("%" PRId64 "\n", all_cost - search(1, K));
	return 0;
}

/*

Nが十分小さいので、以下の方法が使えそう

1. 各区間全部について、そこの両側を変えずにそこを変えたときにコストをどれだけ削減できるかを計算する
   O(N^2 * 計算のコスト)
2. それを用いてDP
   O(N^3)

コストを削減するには、余計な穴を埋めるとよい
→ 両端を見てどっちかに合わせればよい

変更後コスト = 両端のうち塗る行が少なくない方の行数
変更前コスト = ？

止まる行数を加算していく O(N)

*/