Overview

This problem involves a row of \(N\) blocks, where you can freely change the heights of at most \(K\) blocks to minimize the number of “dangerous spots” — places where the height difference between adjacent blocks exceeds \(T\).

By keeping certain blocks unchanged and focusing on the positional and height relationships between them, we can solve this efficiently using Dynamic Programming (DP).

Analysis

1. Freedom in Changing Blocks

Blocks selected for modification can have their heights set to any integer. There are no upper or lower bounds.

2. Focusing on the Gap Between Two Unchanged Blocks

Consider selecting some blocks from left to right that we will not change. Suppose two adjacent “unchanged blocks” are at positions \(p\) and \(i\) (\(p < i\)), and all \(i - p - 1\) blocks between them are changed.

In this case, what is the minimum number of dangerous spots (places where the difference exceeds \(T\)) among the \(i - p\) adjacent pairs from position \(p\) to position \(i\)?

The maximum height change allowed in one step (between adjacent blocks) is \(T\). Therefore, the maximum height difference that can be covered over \(i - p\) steps is \((i - p) \times T\).

• When \(|D_i - D_p| \leq (i - p) \times T\): By setting the heights of the intermediate blocks at appropriately spaced intervals, we can keep all adjacent pair differences within \(T\). Thus, the number of dangerous spots in this segment can be \(0\).
• When \(|D_i - D_p| > (i - p) \times T\): No matter how we set the intermediate heights, at least one pair must have a difference exceeding \(T\). However, by concentrating the large gap into just one spot and keeping the remaining \(i - p - 1\) spots within a difference of \(T\), we can minimize the number of dangerous spots to \(1\).

Thus, the minimum cost (number of dangerous spots) between unchanged blocks \(p\) and \(i\) is simply determined as follows:

\[ \text{cost}(p, i) = \begin{cases} 0 & (\text{if } |D_i - D_p| \leq (i - p) \times T) \\ 1 & (\text{otherwise}) \end{cases} \]

3. Handling the Edges

For the section to the left of the first “unchanged block” and to the right of the last “unchanged block,” since one side’s height is not fixed (can be freely adjusted), the number of dangerous spots can always be made \(0\).

4. Reduction to Dynamic Programming (DP)

From the above analysis, this problem reduces to “selecting some blocks to leave unchanged and minimizing the total cost between the selected blocks.”

Algorithm

We define the following Dynamic Programming (DP):

DP Table Definition

• dp[i][c]: The minimum number of dangerous spots from position \(1\) to \(i\), when the \(i\)-th block is left unchanged and is the \(c\)-th unchanged block counting from the left.

Initialization

When the first unchanged block is at position \(i\), all blocks to its left are changed, so the number of dangerous spots is \(0\). - dp[i][1] = 0 (\(1 \leq i \leq N\)) - All other values are initialized to infinity (\(\infty\)).

Transition

When selecting the \(i\)-th block as the \(c\)-th (\(c \geq 2\)) unchanged block, we transition from the previous unchanged block \(p\) (\(c-1 \leq p < i\)):

\[ \text{dp}[i][c] = \min_{c-1 \leq p < i} \left( \text{dp}[p][c-1] + \text{cost}(p, i) \right) \]

Deriving the Answer

Since the maximum number of blocks that can be changed is \(K\), the number of unchanged blocks \(c\) must be at least \(N - K\) (with a minimum of \(c \geq 1\)).

Therefore, the answer is the minimum value of dp[i][c] over all \(c \geq \max(1, N - K)\), where \(i\) is the last unchanged block. (Since all blocks to the right of the last unchanged block \(i\) are changed, no additional cost is incurred.)

※ When \(K \geq N\), all blocks can be changed, so the answer is unconditionally \(0\).

Complexity

• Time Complexity: \(O(N^3)\) The number of DP states is \(O(N^2)\), and each state requires \(O(N)\) for transitions. Since the constraint is \(N \leq 10\), which is very small, the worst case involves about \(10^3 = 1000\) loop iterations, completing in an instant.
• Space Complexity: \(O(N^2)\) The DP table size is \((N+1) \times (N+1)\), which is well within memory limits.

Implementation Notes

• Index Adjustment: The problem is easier to understand when thought of as 1-indexed (from 1 to \(N\)), so the DP is also constructed with 1-indexing in the code. When accessing values from array D, we convert to 0-indexed with D[i-1] or D[p-1].

• Corner Case Handling: When \(K \geq N\), we output 0 and return early before running the DP, preventing bugs such as out-of-bounds array access.

  Source Code

import sys

def solve():
    # 入力を一括で取得
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    N = int(input_data[0])
    T = int(input_data[1])
    K = int(input_data[2])
    D = [int(x) for x in input_data[3:]]
    
    # すべてのブロックを変更できる場合は危険箇所を 0 にできる
    if K >= N:
        print(0)
        return

    # dp[i][c]: i 番目のブロックを変更せず、そこまでに変更しないブロックを c 個選んだときの最小危険箇所数
    dp = [[float('inf')] * (N + 1) for _ in range(N + 1)]
    
    # 1つ目の変更しないブロックを選ぶ場合、それより左側はすべて変更するためコストは 0
    for i in range(1, N + 1):
        dp[i][1] = 0
        
    # DPの遷移
    for c in range(2, N + 1):
        for i in range(c, N + 1):
            min_val = float('inf')
            for p in range(c - 1, i):
                # p 番目と i 番目のブロックを変更しないときのコスト
                cost = 1 if abs(D[i-1] - D[p-1]) > (i - p) * T else 0
                val = dp[p][c-1] + cost
                if val < min_val:
                    min_val = val
            dp[i][c] = min_val
            
    # 変更しないブロックの数は、少なくとも N - K 個以上必要
    min_c = max(1, N - K)
    ans = float('inf')
    for c in range(min_c, N + 1):
        for i in range(c, N + 1):
            if dp[i][c] < ans:
                ans = dp[i][c]
                
    print(ans)

if __name__ == '__main__':
    solve()

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This editorial was generated by gemini-3.5-flash-thinking.