我如何在无向图中找到桥梁？ [英] How can I find bridges in an undirected graph?

问题描述

给定一个无向图，我怎样才能找到所有的桥梁？我只找到了似乎相当复杂的Tarjan算法。

似乎应该有多个线性时间解决方案，但我找不到任何东西。 b $ b

解决方案

Tarjan的算法是在线性时间运行的无向图中的第一个桥寻找算法。然而，一个更简单的算法存在，你可以看看它的实现 here 。

  private int bridges; //桥梁的数量
 private int cnt; // counter 
 private int [] pre; // pre [v] = dfs检查的顺序v 
 private int [] low; // low [v] =与v 
 
相连的任何顶点的最低预定义公共桥（图G）{
 low = new int [G.V（）]; 
 pre = new int [G.V（）]; 
 for（int v = 0; v  for（int v = 0; v  $ b $ for（int v = 0; v  if（pre [v] == -1）
 dfs（G，v，v ）; 
} 
 
 public int components（）{return bridges + 1; } 
 
 private void dfs（Graph G，int u，int v）{
 pre [v] = cnt ++; 
 low [v] = pre [v]; 
 for（int w：G.adj（v））{
 if（pre [w] == -1）{
 dfs（G，v，w）; 
 low [v] = Math.min（low [v]，low [w]）; 
 if（low [w] == pre [w]）{
 StdOut.println（v + - + w +is a bridge）; 
桥梁++; 
 
 
 
 //更新低位数字 - 忽略导致v 
的边缘反转if（w！= u）
 low [v ] = Math.min（low [v]，pre [w]）; 
 
 
 
 
 $ b $ p 
 $ b 
该算法通过维护2个数组来完成工作，低。 pre保存节点的预序遍历编号。所以pre [0] = 2意味着在第三次dfs调用中发现了顶点0。并且low [u]保存从u到达的任何顶点的最小预序号。 
 
 
算法检测到一个边缘u  -  v，其中u在预编号中首先出现，low [v] == pre [v]。这是因为如果我们移除了u  -  v之间的边，v就无法到达u之前的任何顶点。因此去除边缘会将图分成2个独立的图。  
 
 
有关更详细的解释，您还可以查看这个答案。
 
Given an undirected Graph, how can I find all the bridges? I've only found Tarjan's algorithm which seems rather complicated.

It seems there should be multiple linear time solutions, but I can't find anything.
解决方案
Tarjan's algorithm was the first bridge finding algorithm in an undirected graph that ran in linear time. However a simpler algorithm exists and you can have a look at its implementation here.
    private int bridges;      // number of bridges
    private int cnt;          // counter
    private int[] pre;        // pre[v] = order in which dfs examines v
    private int[] low;        // low[v] = lowest preorder of any vertex connected to v

    public Bridge(Graph G) {
        low = new int[G.V()];
        pre = new int[G.V()];
        for (int v = 0; v < G.V(); v++) low[v] = -1;
        for (int v = 0; v < G.V(); v++) pre[v] = -1;

        for (int v = 0; v < G.V(); v++)
            if (pre[v] == -1)
                dfs(G, v, v);
    }

    public int components() { return bridges + 1; }

    private void dfs(Graph G, int u, int v) {
        pre[v] = cnt++;
        low[v] = pre[v];
        for (int w : G.adj(v)) {
            if (pre[w] == -1) {
                dfs(G, v, w);
                low[v] = Math.min(low[v], low[w]);
                if (low[w] == pre[w]) {
                    StdOut.println(v + "-" + w + " is a bridge");
                    bridges++;
                }
            }

            // update low number - ignore reverse of edge leading to v
            else if (w != u)
                low[v] = Math.min(low[v], pre[w]);
        }
    }
The algorithm does the job by maintaining 2 arrays pre and low. pre holds the pre-order traversal numbering for the nodes. So pre[0] = 2 means that vertex 0 was discovered in the 3rd dfs call. And low[u] holds the smallest pre-order number of any vertex that is reachable from u. 


The algorithm detects a bridge whenever for an edge u--v, where u comes first in the preorder numbering, low[v]==pre[v]. This is because if we remove the edge between u--v, v can't reach any vertex that comes before u. Hence removing the edge would split the graph into 2 separate graphs. 

For a more elaborate explanation you can also have a look at this answer .

                        
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