助教求助：Dynamic Programming Python路径算法作业问题

Hey there! As a TA, I totally get wanting to have a solid, professional grasp on this problem to guide your students—let’s work through this step by step.

First, let’s clarify the problem setup based on your input specs:

• We’ve got a grid where rows represent streets (0 to m, total m+1 rows) and columns represent avenues (0 to n, total n+1 columns).
• Each value in the grid corresponds to a specific (street, avenue) coordinate, e.g., the first line of the m+1 data rows is grid[0][0], grid[0][1], ..., grid[0][n] (all avenues on street 0).

The most common DP path problem for this setup is finding the maximum or minimum sum path from the top-left corner (0,0) to the bottom-right corner (m,n), where you can only move right (to the next avenue) or down (to the next street). I’ll focus on this classic case, but we can adjust if your problem has different movement rules.

Step 1: Define the DP State #

Let’s define dp[i][j] as the optimal sum (max or min) to reach the coordinate (street i, avenue j).

Step 2: State Transition #

To get to (i,j), you can only come from:

• The street above: (i-1, j)
• The avenue to the left: (i, j-1)

So for maximum path sum:

dp[i][j] = grid[i][j] + max(dp[i-1][j], dp[i][j-1])

For minimum path sum, just swap max with min:

dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])

Step 3: Initialize the DP Table #

• The first row (street 0): You can only move right along this row, so each value is the cumulative sum from the start:
  dp[0][j] = dp[0][j-1] + grid[0][j]
• The first column (avenue 0): You can only move down along this column, so each value is the cumulative sum from the start:
  dp[i][0] = dp[i-1][0] + grid[i][0]

Step 4: Full Python Implementation #

Here’s a complete code example that handles input, computes the maximum path sum, and returns the result:

# Read input
m = int(input())
n = int(input())

# Initialize the grid
grid = []
for _ in range(m + 1):
    row = list(map(int, input().split()))
    grid.append(row)

# Create DP table (clear for teaching purposes)
dp = [[0]*(n+1) for _ in range(m+1)]

# Initialize first row
dp[0][0] = grid[0][0]
for j in range(1, n+1):
    dp[0][j] = dp[0][j-1] + grid[0][j]

# Initialize first column
for i in range(1, m+1):
    dp[i][0] = dp[i-1][0] + grid[i][0]

# Fill the rest of the DP table
for i in range(1, m+1):
    for j in range(1, n+1):
        dp[i][j] = grid[i][j] + max(dp[i-1][j], dp[i][j-1])

# The result is the bottom-right corner
print(f"Maximum path sum: {dp[m][n]}")

Step 5: Space Optimization (Optional) #

If you want to save space (since we only need the previous row and current row), we can use a 1D DP array:

# Space-optimized version for maximum path sum
m = int(input())
n = int(input())
grid = [list(map(int, input().split())) for _ in range(m+1)]

dp = [0]*(n+1)
# Initialize first row
dp[0] = grid[0][0]
for j in range(1, n+1):
    dp[j] = dp[j-1] + grid[0][j]

# Fill for remaining rows
for i in range(1, m+1):
    # Update first column of current row
    dp[0] += grid[i][0]
    for j in range(1, n+1):
        dp[j] = grid[i][j] + max(dp[j], dp[j-1])

print(f"Maximum path sum (space-optimized): {dp[n]}")

Notes for Your Students #

• Always start by defining the DP state clearly—this is the most important part of any DP problem.
• Walk through small examples (like m=1, n=1) to verify the logic works.
• If your problem allows different movement (e.g., up/left/diagonal), adjust the state transition to include those possible previous positions.

内容的提问来源于stack exchange，提问作者Justin Haddock