One-way Street Conversion

Converting edges

I’m working through the challenge of directing edges while preserving strong connectivity. Some edges remain two-way to ensure all intersections are reachable.

Analyzing bridge tactics

I'm considering edge orientation to maintain strong connectivity. Identifying and avoiding bridge edges is crucial, as removing them would disrupt connectivity.

Preserving connectivity

Ensuring strong connectivity, especially with bridged edges, is vital. For a directed graph to be strongly connected, it must be "2-edge-connected," meaning no single bridge disconnects the graph.

Piecing together connections

To ensure strong connectivity, only non-bridging edges are directed. Bridges must stay two-way, so first, identify these while building the adjacency list.

Directing alignment

I’m listing the steps: find bridges, label them for two-way orientation, and ensure strong connectivity for non-bridging edges, which can be efficiently directed using DFS or spanning tree methods.

Ensuring connectivity

I'm exploring DFS to maintain Strong connectivity. Each block is directed, forming a directed cycle with no single point of failure. Progressing through connected components to confirm this works.

Ensuring connectivity

OK, let me see. Following the plan, we create adjacency lists, identify bridging edges, and ensure strong connectivity by orienting non-bridging edges using DFS.

Orienting paths

I’m figuring out the strategy: use DFS or BFS to create a spanning tree and label non-bridging edges. Then, I ensure connectivity by checking all nodes.

Ensuring connectivity

I'm piecing together that focusing on 2-edge-connected graphs creates a strong orientation with 2-way bridging edges, boosting the entire graph's connectivity. This approach sufficiently guarantees connectivity.

Testing for accuracy

Crafting the C++ code without comments, I’ll test it against sample input. The result will verify the strong orientation of edge graph connectivity, ensuring bridging edges remain two-way.

Implementing the solution

Crafting the C++ code, I’m reading inputs, building the graph, identifying bridging edges, and conducting a DFS to orient non-bridge edges for strong connectivity while leaving bridging edges two-way.

Mapping the strategy

I’m orienting non-bridging edges with DFS, specifying parent-to-child or child-to-parent orientations, and preserving existing orientations for repeating edges to achieve strong connectivity without using bridging edges.

Thought about traffic conversion using DFS for 21 seconds

#include <bits/stdc++.h>
using namespace std;
int n,m,cases;
vector<vector<int>> graph;
vector<bool> visited,vis2;
vector<int> A,B;
vector<pair<int,int>> edges,bridges;
int timer;

void dfs_bridge(int u,int p){
    visited[u]=true;
    A[u]=B[u]=++timer;
    for(auto &v:graph[u]){
        if(v==p) continue;
        if(visited[v]) B[u]=min(B[u],A[v]);
        else{
            dfs_bridge(v,u);
            B[u]=min(B[u],B[v]);
            if(B[v]>A[u]) bridges.push_back({min(u,v),max(u,v)});
        }
    }
}

vector<vector<int>> adj_no_bridge;
map<pair<int,int>,bool> used;
vector<pair<int,int>> ans;

void dfs_orient(int u){
    vis2[u]=true;
    for(auto &v:adj_no_bridge[u]){
        if(!vis2[v]){
            ans.push_back({u,v});
            dfs_orient(v);
        }
        else if(!used[{u,v}] && !used[{v,u}]){
            ans.push_back({v,u});
        }
        used[{u,v}]=true;
        used[{v,u}]=true;
    }
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cases=0;
    while(true){
        cin>>n>>m;
        if(!cin||(!n&&!m)) break;
        cases++;
        graph.assign(n+1,{});
        edges.clear();
        for(int i=0;i<m;i++){
            int x,y;cin>>x>>y;
            edges.push_back({x,y});
            graph[x].push_back(y);
            graph[y].push_back(x);
        }
        visited.assign(n+1,false);
        A.assign(n+1,0);
        B.assign(n+1,0);
        bridges.clear();
        timer=0;
        dfs_bridge(1,-1);
        for(int i=1;i<=n;i++){
            if(!visited[i]) dfs_bridge(i,-1);
        }
        set<pair<int,int>> is_bridge;
        for(auto &b:bridges) is_bridge.insert(b);
        adj_no_bridge.assign(n+1,{});
        for(auto &e:edges){
            int x=e.first,y=e.second;
            if(is_bridge.count({min(x,y),max(x,y)})) continue;
            adj_no_bridge[x].push_back(y);
            adj_no_bridge[y].push_back(x);
        }
        used.clear();
        ans.clear();
        vis2.assign(n+1,false);
        for(int i=1;i<=n;i++){
            if(!vis2[i]) dfs_orient(i);
        }
        set<pair<int,int>> oriented;
        for(auto &p:ans) oriented.insert(p);
        cout<<cases<<"\n";
        for(auto &e:edges){
            int x=e.first,y=e.second;
            if(is_bridge.count({min(x,y),max(x,y)})){
                cout<<x<<" "<<y<<"\n";
                cout<<y<<" "<<x<<"\n";
            } else {
                if(oriented.count({x,y})) cout<<x<<" "<<y<<"\n";
                else cout<<y<<" "<<x<<"\n";
            }
        }
        cout<<"#\n";
    }
    return 0;
}

Explanation of key ideas (informally, outside the code):

1. Find bridges with a standard DFS (arrays A, B, and a global timer).
2. Any edge that is a bridge cannot be oriented in a single direction if we want a strongly connected final graph, so we output those edges in both directions.
3. All other edges (non-bridges) form 2-edge-connected components. We orient them via a DFS in a way that ensures strong connectivity within each component (tree edges go one way, discovered back edges go the other).
4. Combine by printing each non-bridge edge in one direction, and each bridge edge both ways. That satisfies the requirement that from any intersection, you can reach any other intersection, because bridge edges remain two-way, and the subgraph of non-bridge edges is given a strongly connected orientation.