给定无向图G及MST T：边e的MST归属判断算法设计及结论验证

Hey there! Let's break down these two minimum spanning tree (MST) questions step by step—they're fundamental to understanding how MSTs behave.

Absolutely! We can pull this off using Union-Find (Disjoint Set Union, DSU) for fast connectivity checks, and it fits neatly into the O(V+E) time constraint. Here's the play-by-play:

Let’s define our target edge as e = (u, v) with weight w_e.

• First, construct a subgraph G' that includes every edge in G with a weight strictly less than w_e.
• Next, check if u and v are connected in G':
  • If they’re not connected: There’s no way to link u and v using cheaper edges. Any MST will need to bridge these two disconnected components, so e can definitely be included in some MST (we can pick e instead of any pricier edges that would connect the components).
  • If they are connected: There’s already a path between u and v using edges cheaper than e. Adding e to any spanning tree would make the total weight larger than necessary, so e can’t be part of any MST.

Union-Find operations are nearly constant time (amortized), and building the subgraph plus running the connectivity check all adds up to O(V+E) time—exactly what we need.

First, let’s restate the conclusion clearly to avoid confusion:

If adding edge e to an MST T creates a cycle, and there’s an edge in that cycle (other than e) with a weight less than e, then e cannot be part of any MST.

This conclusion is completely valid. Here’s the reasoning:
Since T is an MST, every edge in the cycle formed by adding e must have a weight ≤ w_e—if there was an edge in T heavier than e, we could swap that edge with e to get a smaller total weight, which would contradict T being an MST. Now, if there’s an edge in the cycle with weight strictly less than w_e, that means u and v were already connected via cheaper edges (the path in T that includes this lighter edge). As we saw in Question 1, this means e can’t be part of any MST.

A quick side note: If the cycle only has edges with weight equal to w_e, then e can be part of another MST—we can swap e with any of the equal-weight edges in the cycle to get a different MST with the same total weight. But that’s a separate case that doesn’t break the original conclusion.

If we needed to design the check without relying on an existing MST T (since the conclusion ties to a specific T), we’d just use the O(V+E) algorithm from Question 1—it’s more general because it doesn’t require a precomputed MST to start with.

内容的提问来源于stack exchange，提问作者Vishavjeet Singh