Overview

In this problem, we need to paint \(N\) planks using \(K\) colors, minimizing the total cost under the constraint that adjacent planks must not have the same color. While this is a standard linear DP (DP along a sequence), a naive transition is too slow. Therefore, we speed up the transitions by using the minimum and second minimum values.

Analysis

DP Definition

It is natural to determine the colors from left to right. Let

\[\mathrm{dp}[i][j]\]

be the minimum cost to paint planks \(1\) through \(i\) such that plank \(i\) is painted with color \(j\). Then, the transition is:

\[\mathrm{dp}[i][j] = C_{i,j} + \min_{j' \neq j} \mathrm{dp}[i-1][j']\]

(plank \(i-1\) can be painted with any color \(j'\) other than \(j\)). The base cases are \(\mathrm{dp}[1][j] = C_{1,j}\), and the answer is \(\min_j \mathrm{dp}[N][j]\).

Why the Naive Implementation Fails

If we implement the above formula directly, for each \(i\) and each \(j\), computing \(\min_{j'\neq j}\) takes \(O(K)\) time, resulting in an overall time complexity of \(O(N K^2)\).

The constraint is \(N \times K \leq 10^6\), which means cases such as \(N = 1, K = 10^6\) or \(N = 1000, K = 1000\) can occur. When \(N = 1000, K = 1000\), \(NK^2 = 10^9\), which will definitely result in TLE.

Idea for Optimization

The term \(\min_{j' \neq j} \mathrm{dp}[i-1][j']\) represents “the minimum value excluding \(j\)”. The key observation here is:

• Let \(i_1\) be the index of the minimum value across the entire array, \(m_1\) be that minimum value, and \(m_2\) be the minimum value excluding \(i_1\) (i.e., the second smallest value). Then:
  • When \(j \neq i_1\), the position of the minimum value \(i_1\) is not excluded, so \(\min_{j'\neq j} = m_1\).
  • Only when \(j = i_1\), \(\min_{j'\neq j} = m_2\).

In other words, “the minimum value excluding \(j\)” has only \(2\) possible values depending on \(j\). Therefore,

\[\mathrm{dp}[i][j] = \begin{cases} C_{i,j} + m_2 & (j = i_1) \\ C_{i,j} + m_1 & (j \neq i_1)\end{cases}\]

This allows us to compute the transitions for one entire step in \(O(K)\) time.

Concrete Example

When \(K=4\) and \(\mathrm{dp}[i-1] = [7, 3, 9, 5]\), we have \(m_1 = 3\) (at position \(i_1 = 2\), 1-indexed) and \(m_2 = 5\). If we paint plank \(i\) with any color other than color \(2\), the previous plank can be painted with color \(2\), adding a cost of \(+3\). Only when painting plank \(i\) with color \(2\), the added cost is \(+5\).

Algorithm

1. Initialize \(\mathrm{dp} = (C_{1,1}, \dots, C_{1,K})\).
2. For \(i = 2, 3, \dots, N\), do the following:
   • Find the minimum value \(m_1\) of \(\mathrm{dp}\) and its position \(i_1\), as well as the minimum value \(m_2\) excluding \(i_1\).
   • Construct the new array \(\mathrm{ndp}\) as \(\mathrm{ndp}[j] = C_{i,j} + m_1\) (except for \(\mathrm{ndp}[i_1] = C_{i,i_1} + m_2\)).
   • Update \(\mathrm{dp} \leftarrow \mathrm{ndp}\).
3. Output \(\min_j \mathrm{dp}[j]\).

Note that since \(K \geq 2\) is guaranteed, \(m_2\) always exists (if \(K \ge 2\), there is at least one element other than \(i_1\)), and a valid coloring is always possible.

Complexity

• Time complexity: \(O(NK)\) (sufficiently fast since \(NK \leq 10^6\))
• Space complexity: \(O(K)\) (we only need to keep one row of the \(\mathrm{dp}\) array; \(O(NK)\) if including the storage for the input)

Implementation Notes

• Case \(N = 1\): Since no transitions occur, simply output \(\min_j C_{1,j}\) (writing the logic so that it is handled solely by the loop initialization helps prevent bugs). In the code above, it is explicitly handled with a conditional branch.

• Finding the second minimum: After finding the index of the minimum value \(i_1\), take the smaller of min(dp[:i1]) and min(dp[i1+1:]). When \(i_1\) is at an edge (\(0\) or \(K-1\)), one side will be empty, so handle these cases to avoid min() raising an error on an empty list.

  • Of course, an implementation that “finds the minimum and second minimum simultaneously in a single loop” is also fine, but in Python, using built-in functions implemented in C (like min and index) is faster.

• Fast I/O: Since we read up to \(NK \leq 10^6\) integers, instead of calling input() repeatedly in Python, read everything at once using sys.stdin.buffer.read().split() and convert to int only when necessary (in this code, applying map(int, ...) when processing each row avoids unnecessary conversions and large intermediate lists).

• Integer Overflow: The total cost can reach up to around \(10^9 \times 10^5 = 10^{14}\), so languages like C++ require long long (not an issue in Python due to arbitrary-precision integers).

• Using a list comprehension such as [v + m1 for v in vals] to add in bulk, and then overwriting only the position of the minimum value afterwards, is both concise and fast.

  Source Code

import sys

def main():
    data = sys.stdin.buffer.read().split()
    n = int(data[0]); k = int(data[1])
    pos = 2
    dp = list(map(int, data[pos:pos + k]))
    pos += k

    if n == 1:
        sys.stdout.write(str(min(dp)) + "\n")
        return

    for _ in range(n - 1):
        m1 = min(dp)
        i1 = dp.index(m1)
        if i1 == 0:
            m2 = min(dp[1:])
        elif i1 == k - 1:
            m2 = min(dp[:k - 1])
        else:
            a = min(dp[:i1])
            b = min(dp[i1 + 1:])
            m2 = a if a < b else b

        vals = list(map(int, data[pos:pos + k]))
        pos += k
        ndp = [v + m1 for v in vals]
        ndp[i1] = vals[i1] + m2
        dp = ndp

    sys.stdout.write(str(min(dp)) + "\n")

main()

---

This editorial was generated by claude5opus-high.