Overview

Given a corridor of \(N\) tiles, you move from tile \(1\) to tile \(N\) by taking steps of 1 or 2 tiles at a time. The goal is to minimize the number of tiles where footprints (traces of dust reduction) are left. This can be solved using dynamic programming (DP).

Analysis

Conditions for Leaving Footprints

When stepping on tile \(i\), a footprint is left when the dust thickness actually decreases. This happens when both of the following conditions are satisfied simultaneously:

• \(H_i > 0\) (there is dust originally)
• \(D_i > 0\) (stepping reduces the dust)

Conversely, if \(H_i = 0\) (there is no dust to begin with) or \(D_i = 0\) (stepping does not reduce the dust), no footprint is left even when stepping on the tile.

Movement Options

When on tile \(j\), you can move to tile \(j+1\) or tile \(j+2\). In other words, you can skip a tile by jumping over one. The idea is to skip tiles that would leave footprints as much as possible.

Reduction to DP

Assign a “footprint cost” of \(0\) or \(1\) to each tile based on whether stepping on it leaves a footprint. The problem becomes minimizing the total cost of tiles stepped on along a path from tile \(1\) to tile \(N\).

This is a classic 1D DP problem (the so-called “frog jump” problem).

Algorithm

1. For each tile \(i\), compute the footprint cost \(\text{cost}[i]\).

   • If \(H_i > 0\) and \(D_i > 0\), then \(\text{cost}[i] = 1\); otherwise \(\text{cost}[i] = 0\).

2. Define the DP array \(\text{dp}[i]\) as “the minimum number of footprints to reach tile \(i\).”

3. Initial value: \(\text{dp}[0] = \text{cost}[0]\) (tile \(1\) must always be stepped on).

4. Transitions:

   • \(\text{dp}[1] = \text{dp}[0] + \text{cost}[1]\) (tile \(2\) can only be reached in one step)
   • For \(i \geq 2\): \(\text{dp}[i] = \min(\text{dp}[i-1],\ \text{dp}[i-2]) + \text{cost}[i]\)

5. The answer is \(\text{dp}[N-1]\).

Concrete Example

For example, with \(N = 5\), \(H = [1, 1, 0, 1, 1]\), \(D = [1, 1, 1, 1, 1]\):

• \(\text{cost} = [1, 1, 0, 1, 1]\) (tile 3 has no dust, so its cost is 0)
• Path \(1 \to 2 \to 3 \to 5\) (skipping tile 4) gives footprint count \(1+1+0+1 = 3\)
• Path \(1 \to 3 \to 5\) (skipping tiles 2 and 4) gives footprint count \(1+0+1 = 2\)

The DP automatically finds the optimal path.

Complexity

• Time complexity: \(O(N)\) — constant-time processing for each tile
• Space complexity: \(O(N)\) — for the DP array and cost array

Implementation Notes

• Be careful not to confuse 0-indexed and 1-indexed. The problem statement uses 1-indexed, but the code processes in 0-indexed.

• When \(H_i = 0\), the dust does not decrease regardless of the value of \(D_i\) (due to the rule that it cannot go below \(0\)), so no footprint is left. It is important to correctly check this condition.

• When \(N = 2\), you have no choice but to go directly from tile \(1\) to tile \(2\) in one step with no room for skipping; this edge case is also handled correctly.

  Source Code

import sys

def solve():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    H = [int(input_data[idx + i]) for i in range(N)]; idx += N
    D = [int(input_data[idx + i]) for i in range(N)]; idx += N
    
    # foot[i] = 1 if stepping on tile i leaves a footprint (H[i] > 0 and D[i] > 0)
    # If H[i] == 0 or D[i] == 0, stepping on it leaves no footprint
    # We want to go from tile 0 to tile N-1 (0-indexed), stepping on each visited tile
    # Minimize the number of tiles where footprint is left
    
    # cost[i] = 1 if visiting tile i leaves a footprint, 0 otherwise
    # A footprint is left when the dust decreases, i.e., H[i] > 0 and D[i] > 0
    cost = [0] * N
    for i in range(N):
        if H[i] > 0 and D[i] > 0:
            cost[i] = 1
    
    # DP: dp[i] = minimum footprints to reach tile i
    # Transitions: from tile i, go to i+1 or i+2
    # We must start at tile 0 and end at tile N-1
    
    INF = float('inf')
    dp = [INF] * N
    dp[0] = cost[0]
    
    if N >= 2:
        dp[1] = dp[0] + cost[1]
    
    for i in range(2, N):
        dp[i] = min(dp[i-1], dp[i-2]) + cost[i]
    
    print(dp[N-1])

solve()

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This editorial was generated by claude4.6opus-thinking.