Slimes

Problem Overview

Learning Goals

After solving this problem, you will be able to:

• Recognize interval DP problems where merging/splitting is involved
• Define DP states over intervals [i, j] instead of single indices
• Use prefix sums to compute range sums in O(1)
• Iterate by interval length to ensure subproblems are solved first

Problem Statement

Problem: There are N slimes arranged in a row. The i-th slime has size a[i]. You can merge two adjacent slimes into one. When slimes of size x and y merge, the resulting slime has size x+y and the cost of this operation is x+y. Find the minimum total cost to merge all slimes into a single slime.

Input:

• Line 1: N (number of slimes)
• Line 2: a[1], a[2], …, a[N] (sizes of slimes)

Output:

• Minimum total cost to merge all slimes

Constraints:

• 2 <= N <= 400
• 1 <= a[i] <= 10^9

Example

Input:
4
10 20 30 40

Output:
190

Explanation:

• Merge slimes 1 and 2: cost = 10 + 20 = 30, result = [30, 30, 40]
• Merge slimes 2 and 3: cost = 30 + 40 = 70, result = [30, 70]
• Merge slimes 1 and 2: cost = 30 + 70 = 100, result = [100]
• Total cost = 30 + 70 + 100 = 190 (but actually optimal is different, see dry run)

Intuition: How to Think About This Problem

Pattern Recognition

Key Question: Why does the merge order matter, and how do we find the optimal order?

This is a classic Interval DP problem. The key insight is that to merge all slimes in an interval [i, j], we must at some point have exactly two groups: one covering [i, k] and one covering [k+1, j]. The cost of the final merge is the sum of all elements in [i, j], which is fixed regardless of how we formed the two groups. Thus, we recursively minimize the cost of forming each group.

Breaking Down the Problem

1. What are we looking for? Minimum total cost to merge all slimes into one.
2. What information do we have? The sizes of all slimes and adjacency constraint.
3. What is the key observation? When merging interval [i,j], the final merge always costs sum(a[i]…a[j]), but the internal costs depend on the split point.

Analogies

Think of this problem like merging sorted files or optimal binary search tree construction. Each merge combines two groups into one, and we pay the “weight” of everything involved. The goal is to find the merge tree with minimum total weight.

Another analogy: imagine you have N piles of stones. Each time you merge two adjacent piles, you pay a cost equal to the total stones in both piles. This is identical to the slimes problem.

Solution 1: Recursive with Memoization

Idea

Try all possible ways to split interval [i, j] into two parts, recursively solve each part, and add the merge cost (sum of all elements in [i, j]).

Algorithm

1. For each interval [i, j], try all split points k where i <= k < j
2. Recursively compute cost for [i, k] and [k+1, j]
3. Add prefix[j+1] - prefix[i] as the merge cost
4. Memoize results to avoid recomputation

Code

import sys
from functools import lru_cache

def solve_recursive(n: int, sizes: list[int]) -> int:
  """
  Recursive solution with memoization.

  Time: O(N^3)
  Space: O(N^2)
  """
  # Compute prefix sums for O(1) range sum queries
  prefix = [0] * (n + 1)
  for i in range(n):
    prefix[i + 1] = prefix[i] + sizes[i]

  def range_sum(i: int, j: int) -> int:
    return prefix[j + 1] - prefix[i]

  @lru_cache(maxsize=None)
  def dp(i: int, j: int) -> int:
    # Base case: single slime needs no merging
    if i == j:
      return 0

    # Try all split points
    result = float('inf')
    for k in range(i, j):
      cost = dp(i, k) + dp(k + 1, j) + range_sum(i, j)
      result = min(result, cost)
    return result

  return dp(0, n - 1)

# Read input and solve
if __name__ == "__main__":
  n = int(input())
  sizes = list(map(int, input().split()))
  print(solve_recursive(n, sizes))

Complexity

Why This Works (But Can Be Slower in Practice)

The recursive approach has function call overhead and may hit Python’s recursion limit for large inputs. The iterative approach below is generally preferred for competitive programming.

Solution 2: Iterative Interval DP (Optimal)

Key Insight

The Trick: Process intervals in order of increasing length. This ensures that when computing dp[i][j], all shorter intervals dp[i][k] and dp[k+1][j] have already been computed.

DP State Definition

In plain English: dp[i][j] answers “What is the cheapest way to combine slimes i through j into one?”

State Transition

dp[i][j] = min(dp[i][k] + dp[k+1][j]) + sum(a[i]...a[j])  for all k in [i, j-1]

Why? To merge [i, j] into one slime:

1. First merge [i, k] into one slime (cost: dp[i][k])
2. Then merge [k+1, j] into one slime (cost: dp[k+1][j])
3. Finally merge these two slimes (cost: sum of all elements = prefix[j+1] - prefix[i])

Base Cases

Algorithm

1. Compute prefix sums for O(1) range queries
2. Initialize dp[i][i] = 0 for all i
3. Iterate by interval length from 2 to N
4. For each interval [i, j], try all split points and take minimum
5. Return dp[0][N-1]

Dry Run Example

Let’s trace through with input N = 4, sizes = [10, 20, 30, 40]:

Prefix sums: [0, 10, 30, 60, 100]
  - sum(0,0) = 10, sum(0,1) = 30, sum(0,2) = 60, sum(0,3) = 100
  - sum(1,1) = 20, sum(1,2) = 50, sum(1,3) = 90
  - sum(2,2) = 30, sum(2,3) = 70
  - sum(3,3) = 40

Initial dp (single slimes):
  dp[0][0] = 0, dp[1][1] = 0, dp[2][2] = 0, dp[3][3] = 0

Length 2 intervals:
  dp[0][1]: split at k=0
    cost = dp[0][0] + dp[1][1] + sum(0,1) = 0 + 0 + 30 = 30
    dp[0][1] = 30

  dp[1][2]: split at k=1
    cost = dp[1][1] + dp[2][2] + sum(1,2) = 0 + 0 + 50 = 50
    dp[1][2] = 50

  dp[2][3]: split at k=2
    cost = dp[2][2] + dp[3][3] + sum(2,3) = 0 + 0 + 70 = 70
    dp[2][3] = 70

Length 3 intervals:
  dp[0][2]: try k=0 and k=1
    k=0: dp[0][0] + dp[1][2] + sum(0,2) = 0 + 50 + 60 = 110
    k=1: dp[0][1] + dp[2][2] + sum(0,2) = 30 + 0 + 60 = 90
    dp[0][2] = min(110, 90) = 90

  dp[1][3]: try k=1 and k=2
    k=1: dp[1][1] + dp[2][3] + sum(1,3) = 0 + 70 + 90 = 160
    k=2: dp[1][2] + dp[3][3] + sum(1,3) = 50 + 0 + 90 = 140
    dp[1][3] = min(160, 140) = 140

Length 4 interval (final answer):
  dp[0][3]: try k=0, k=1, k=2
    k=0: dp[0][0] + dp[1][3] + sum(0,3) = 0 + 140 + 100 = 240
    k=1: dp[0][1] + dp[2][3] + sum(0,3) = 30 + 70 + 100 = 200
    k=2: dp[0][2] + dp[3][3] + sum(0,3) = 90 + 0 + 100 = 190
    dp[0][3] = min(240, 200, 190) = 190

Answer: 190

Visual Diagram

Slimes: [10, 20, 30, 40]
         0   1   2   3

Optimal merge order (k=2 for full interval, k=1 for [0,2]):

Step 1: Merge [0,1] -> cost 30
  [10, 20, 30, 40] -> [30*, 30, 40]

Step 2: Merge [30*, 30] -> cost 60
  [30*, 30, 40] -> [60*, 40]

Step 3: Merge [60*, 40] -> cost 100
  [60*, 40] -> [100*]

Total: 30 + 60 + 100 = 190

Code

Python Solution:

def solve(n: int, sizes: list[int]) -> int:
  """
  Iterative interval DP solution for Slimes.

  Time: O(N^3)
  Space: O(N^2)
  """
  # Prefix sums for O(1) range sum queries
  prefix = [0] * (n + 1)
  for i in range(n):
    prefix[i + 1] = prefix[i] + sizes[i]

  # dp[i][j] = minimum cost to merge slimes [i, j]
  INF = float('inf')
  dp = [[INF] * n for _ in range(n)]

  # Base case: single slimes cost nothing
  for i in range(n):
    dp[i][i] = 0

  # Fill DP by increasing interval length
  for length in range(2, n + 1):
    for i in range(n - length + 1):
      j = i + length - 1
      range_sum = prefix[j + 1] - prefix[i]

      # Try all split points
      for k in range(i, j):
        cost = dp[i][k] + dp[k + 1][j] + range_sum
        dp[i][j] = min(dp[i][j], cost)

  return dp[0][n - 1]


if __name__ == "__main__":
  n = int(input())
  sizes = list(map(int, input().split()))
  print(solve(n, sizes))

Complexity

Common Mistakes

Mistake 1: Wrong Iteration Order

# WRONG: Iterating by start/end instead of length
for i in range(n):
  for j in range(i, n):
    for k in range(i, j):
      dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + ...)

Problem: When computing dp[i][j], the subproblems dp[i][k] and dp[k+1][j] may not have been computed yet.

Fix: Iterate by interval length from smallest to largest:

for length in range(2, n + 1):
  for i in range(n - length + 1):
    j = i + length - 1
    # Now all shorter intervals are already computed

Mistake 2: Integer Overflow

Problem: With a[i] up to 10^9 and N up to 400, the total cost can exceed 10^11.

Fix: Use long long in C++ or Python’s native integers:

Mistake 3: Forgetting to Add Merge Cost

# WRONG: Only counting recursive costs
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j])  # Missing merge cost!

Problem: The cost of merging the two resulting slimes is not counted.

Fix: Always add the range sum as merge cost:

dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + range_sum(i, j))

Edge Cases

When to Use This Pattern

Use Interval DP When:

• You need to merge/split adjacent elements optimally
• The problem involves contiguous subarrays/substrings
• Optimal substructure exists over intervals [i, j]
• The answer for [i, j] depends on answers for sub-intervals

Classic Interval DP Problems:

• Matrix chain multiplication
• Optimal binary search tree
• Burst balloons
• Palindrome partitioning
• Stone game (merge stones)

Don’t Use When:

• Elements can be processed in any order (not just adjacent)
• The problem doesn’t have optimal substructure over intervals
• A greedy approach works (verify carefully!)

Pattern Recognition Checklist:

• Are we combining/splitting adjacent elements? -> Consider Interval DP
• Is there a cost associated with each merge/split? -> Classic Interval DP
• Does order of operations matter? -> Need to try all orderings
• Can the problem be divided at a “split point”? -> Interval DP likely applies

Related Problems

Easier (Do These First)

Similar Difficulty

Harder (Do These After)

Key Takeaways

1. The Core Idea: Interval DP solves problems where we need to find the optimal way to process contiguous segments by trying all possible split points.

2. Time Optimization: The key is iterating by interval length to ensure all subproblems are solved before they are needed.

3. Space Trade-off: We use O(N^2) space for the DP table to avoid recomputation of overlapping subproblems.

4. Pattern: This belongs to the “Interval DP” family, which is essential for merge/split optimization problems.

Practice Checklist

Before moving on, make sure you can:

• Solve this problem without looking at the solution
• Explain why we iterate by interval length
• Derive the recurrence relation from scratch
• Implement both recursive and iterative versions
• Identify interval DP patterns in new problems

Additional Resources

• AtCoder DP Contest - Problem N - Original problem
• CP-Algorithms: Interval DP
• Knuth’s Optimization - Advanced optimization for interval DP