问题
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.
回答1:
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 .
来源：https://stackoverflow.com/questions/28917290/how-can-i-find-bridges-in-an-undirected-graph