From d71651b14f3dc57c451ff7f467459a38770ecc7e Mon Sep 17 00:00:00 2001 From: fffelix-huang Date: Sat, 16 Sep 2023 22:08:10 +0800 Subject: [PATCH] Fix typos --- src/graph/bridge-searching-online.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/src/graph/bridge-searching-online.md b/src/graph/bridge-searching-online.md index da3af4402..702dc5a88 100644 --- a/src/graph/bridge-searching-online.md +++ b/src/graph/bridge-searching-online.md @@ -79,7 +79,7 @@ We will now consistently disassemble every operation that we need to learn to im For example you can re-root the tree of vertex $a$, and then attach it to another tree by setting the ancestor of $a$ to $b$. However the question about the effectiveness of the re-rooting operation arises: - in order to re-root the tree with the root $r$ to the vertex $v$, it is necessary to necessary to visit all vertices on the path between $v$ and $r$ and redirect the pointers `par[]` in the opposite direction, and also change the references to the ancestors in the DSU that is responsible for the connected components. + in order to re-root the tree with the root $r$ to the vertex $v$, it is necessary to visit all vertices on the path between $v$ and $r$ and redirect the pointers `par[]` in the opposite direction, and also change the references to the ancestors in the DSU that is responsible for the connected components. Thus, the cost of re-rooting is $O(h)$, where $h$ is the height of the tree. You can make an even worse estimate by saying that the cost is $O(\text{size})$ where $\text{size}$ is the number of vertices in the tree. @@ -103,7 +103,7 @@ We will now consistently disassemble every operation that we need to learn to im * Searching for the cycle formed by adding a new edge $(a, b)$. Since $a$ and $b$ are already connected in the tree we need to find the [Lowest Common Ancestor](lca.md) of the vertices $a$ and $b$. - The cycle will consist of the paths from $b$ to the LCA, from the LCA to $b$ and the edge $a$ to $b$. + The cycle will consist of the paths from $b$ to the LCA, from the LCA to $a$ and the edge $a$ to $b$. After finding the cycle we compress all vertices of the detected cycle into one vertex. This means that we already have a complexity proportional to the cycle length, which means that we also can use any LCA algorithm proportional to the length, and don't have to use any fast one. pFad - Phonifier reborn

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