HackerRank《Repairing Roads》谜题技术求解咨询

Alright, let's break down how to solve this problem efficiently—this is a classic graph theory problem in disguise, so once you spot the pattern, it's straightforward.

Understanding the Problem Core #

First, let's translate the robot's behavior into graph terms:

• Each city is a vertex, each road is an undirected edge.
• A robot's repair path is an edge-disjoint trail (it moves along unvisited edges, repairing them one by one until no adjacent unvisited edges are left).

Our goal is to find the minimum number of such trails needed to cover every edge in the connected graph (since the problem states all cities are reachable from each other).

Key Graph Theory Insight #

This problem directly maps to finding the minimum number of trails needed to cover all edges of a connected undirected graph. The solution comes from Eulerian trail properties:

• If every vertex has an even degree (number of roads connected to it), we can use 1 robot—this is an Eulerian circuit, where the robot can start at any vertex, traverse every edge exactly once, and end back at the starting point.
• If there are 2k vertices with odd degrees (undirected graphs always have an even number of odd-degree vertices), we need k robots. Each robot will traverse a trail connecting two of these odd-degree vertices, and together these trails cover all edges.

Step-by-Step Solution #

1. Calculate Vertex Degrees: For each city, count how many roads are connected to it.
2. Count Odd-Degree Vertices: Tally up how many cities have an odd number of connected roads.
3. Compute Result: The minimum number of robots is max(1, odd_count // 2). We use max(1, ...) because even if all degrees are even (odd_count=0), we still need at least one robot to do the work.

Example Walkthroughs #

• Triangle Graph (3 cities, 3 roads): Each city has degree 2 (even). Odd count is 0 → 1 robot suffices.
• Linear Path (3 cities, 2 roads): Two end cities have degree 1 (odd), middle has degree 2. Odd count is 2 → 1 robot.
• Cross Shape (5 cities, 4 roads): Four outer cities have degree 1 (odd), center has degree 4. Odd count is 4 → 2 robots.

Code Implementation (Python) #

Here's a concise implementation that follows the logic above:

n, m = map(int, input().split())
degree = [0] * (n + 1)  # Assuming cities are numbered from 1 to n

for _ in range(m):
    u, v = map(int, input().split())
    degree[u] += 1
    degree[v] += 1

odd_count = sum(1 for d in degree if d % 2 != 0)
print(max(1, odd_count // 2))

Notes #

• The problem guarantees the graph is connected, so we don't need to handle multiple disconnected components (if it were disconnected, we'd calculate the result for each component and sum them up).
• Edge cases like m=0 (no roads to repair) aren't part of the problem's scope since it states all roads need repair—but the code would still output 1, which is harmless if such a case isn't present in test data.

内容的提问来源于stack exchange，提问作者Ronbear