I have this code to find bridges in a connected graph:
void dfs (int v, int p = -1) {
used[v] = true;
tin[v] = fup[v] = timer++;
for (size_t i=0; i<g[v].size(); ++i) {
int to = g[v][i];
if (to == p) continue;
if (used[to])
fup[v] = min (fup[v], tin[to]);
else {
dfs (to, v);
fup[v] = min (fup[v], fup[to]);
if (fup[to] > tin[v])
printf("%d %d", v, to);
}
}
}
How to rewrite it without using recursion? I know, it's possible to do it and I should use stack, but this line must be executed after recursive call of dfs() and I can't achieve with a stack:
fup[v] = min(fup[v], fup[to])
So, how to rewrite my algorithm iteratively?
Find Bridges in a graph using DFS traversal Do DFS traversal of the given graph. In DFS tree an edge (u, v) (u is parent of v in DFS tree) is bridge if there does not exist any other alternative edge to reach u or an ancestor of u from subtree rooted with v.
In graph theory, a bridge, isthmus, cut-edge, or cut arc is an edge of a graph whose deletion increases the graph's number of connected components. Equivalently, an edge is a bridge if and only if it is not contained in any cycle.
If a graph is not connected, we can increase the number of bridges by connecting two connected components by an edge. Thus a graph with the maximal number of bridges is connected. In a connected graph, the bridges form a tree connecting the connected components that would be left if all the bridges were removed.
You want to make a "stack frame" structure
struct Frame {
Frame(int v, int p, int i, Label label);
int v;
int p;
int i;
};
// constructor here
and, as you say, a stack<Frame>
. Between all of these fields, it's possible to simulate the call stack (untested code to give the general idea).
void dfs(int v, int p = -1) {
stack<Frame> st;
st.push(Frame(v, p, 0));
do {
Frame fr(st.top());
st.pop();
v = fr.v;
p = fr.p;
int i(fr.i);
if (i > 0) {
int to(g[v][i - 1]);
fup[v] = min(fup[v], fup[to]);
if (fup[to] > tin[v]) { printf("%d %d", v, to); }
if (i == g[v].size()) { continue; }
} else if (i == 0) {
used[v] = true;
tin[v] = fup[v] = timer++;
}
int to(g[v][i]);
if (to == p) { continue; }
if (used[to]) {
fup[v] = min(fup[v], tin[to]);
} else {
st.push(Frame(to, v, 0));
}
st.push(Frame(v, p, i + 1));
} while (!st.empty());
}
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