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But here is a simple way using the DFS tree: give each back-edge an unique index starting from $$$N + 1$$$; for each vertex $$$u$$$, calculate the index of the back-edge $$$u$$$ is under; call that $$$\mathrm {cycleId} [u]$$$; if $$$u$$$ isn't in a cycle then $$$\mathrm {cycleId} [u] = u$$$;
3 lip 2019 · The DFS tree and observation 1 are the core of Tarjan's bridge-finding algorithm. Typical tutorials about finding bridges only mention the DFS tree in passing and start by defining weird arrays like $$$\mathrm{dfs}[u]$$$ and $$$\mathrm{low}[u]$$$: forget all that.
We all know of various problems using DP like subset sum, knapsack, coin change etc. We can also use DP on trees to solve some specific problems. We define functions for nodes of the trees, which we calculate recursively based on children of a nodes.
This is an introductory tutorial that discu...
15 lut 2024 · Unlike linear data structures (Array, Linked List, Queues, Stacks, etc) which have only one logical way to traverse them, trees can be traversed in different ways. Generally, there are 2 widely used ways for traversing trees: DFS or Depth-First Search. BFS or Breadth-First Search.
The idea of a bridge tree is to shrink the maximal components without a bridge into one node, leaving only the bridges of the original graph as the edges in the bridge tree. In the above image, the bridges of the graph are marked in red.
Given a tree, every vertex has a color. The query is how many vertices in the subtree of vertex v are colored with color c? First, calculate the size of the subtree of every vertex, it can be done with a single dfs $$$(sz[v] = v + all(sz[child \ of \ v]))$$$
