One-way Street Conversion

Mapping out strong connectivity

I’m analyzing the problem of converting two-way streets into one-way streets, ensuring each intersection is reachable from every other intersection. This involves maintaining strong connectivity in the directed graph, making key decisions on necessary streets.

Navigating bridging gaps

I'm piecing together how to handle edges that can't be directed due to the presence of a bridge, which disrupts strong connectivity in the undirected graph.

Ensuring connectivity

To maintain strong connectivity, each test case identifies all bridges. Direct non-bridge edges, focusing on DFS for orienting edges in bridgeless components, particularly in connected graphs.

Mapping the approach

I’m planning to read the graph parameters, build an adjacency list, identify bridges using a DFS algorithm, and then direct edges to maintain strong connectivity, ensuring non-bridge edges are oriented.

Ensuring edge orientation

Directing edges consistently ensures strong connectivity, outputting non-bridge edges in a single direction and displaying bridge edges bi-directionally. The solution will be encoded in C++.

Three-task plan

First, clear data structures. Then, create adjacency lists for each node and identify bridges with DFS. Finally, orient non-bridge edges using DFS or BFS from node 1.

Navigating edge orientations

I'm thinking through directing non-bridge edges consistently by comparing visit times, ensuring coverage and strong connectivity in the graph.

Ensuring consistency

I’m focusing on a DFS to orient edges from parent to child for non-bridge edges, ensuring consistency, and printing bridge edges in both directions. Coding can now begin.

Thought about one-way street conversion for 42 seconds

Below is an illustrative C++ solution that uses a bridge-finding routine and then orients each edge accordingly to ensure (as much as possible) that the resulting directed graph remains strongly connected.

Key observations and approach:

1. An undirected graph can be oriented to be strongly connected if and only if the graph is 2-edge-connected (i.e., it has no bridges).
2. Therefore, any edge that is a bridge in the undirected graph must be printed in both directions (because orienting a bridge in just one direction will break strong connectivity).
3. All non-bridge edges (those that lie on cycles) can be oriented in a single direction. A common approach is to do a DFS and orient each edge from the DFS tree parent to child, and for any back edge, also orient from the current node to the ancestor.

Below is a skeleton code that demonstrates the idea. It uses:

• A DFS-based bridge-finding routine (the one you indicated).
• Another DFS to orient the non-bridge edges.
• For each test case, we print the result as specified.

You can certainly optimize or adapt details (for instance, you might store the final orientation during the same DFS you use for bridge-finding, but here we keep it conceptually clearer by separating them).

#include <bits/stdc++.h>
using namespace std;

// We assume n up to 1000
static const int MAXN = 1000;

// Global or static-like structures for each test case.
// You could encapsulate these in a class if preferred.
vector<int> graph[MAXN+1];
bool visited[MAXN+1];
int A[MAXN+1], B[MAXN+1]; // Times of discovery (A) and low-link values (B)
int timer;
vector<pair<int,int>> bridges;

// We will store edges in a separate list as well to print them later
//   and to know which are bridges easily.
vector<pair<int,int>> edges; 

// We can use a set (or unordered_set) of pairs to mark bridges.
set<pair<int,int>> is_bridge;

int n, m; // number of intersections, number of streets

// Standard DFS for bridge-finding.
void dfs_bridge(int node, int parent) {
    visited[node] = true;
    A[node] = B[node] = timer++;
    for (int child : graph[node]) {
        if (child == parent) continue;
        if (visited[child]) {
            // Back edge
            B[node] = min(B[node], A[child]);
        } else {
            dfs_bridge(child, node);
            B[node] = min(B[node], B[child]);
            // Condition for a bridge
            if (B[child] > A[node]) {
                // Mark it as a bridge in both directions
                is_bridge.insert({min(node, child), max(node, child)});
            }
        }
    }
}

// We will also do a DFS to orient non-bridge edges
// We'll store directed edges in a vector of pairs (u -> v).
vector<pair<int,int>> directed_edges;

// We need a visited array for orientation DFS distinct from the one for bridges
bool used_for_orientation[MAXN+1];

void orient_dfs(int node, int parent) {
    used_for_orientation[node] = true;
    for (int child : graph[node]) {
        // If this is the edge to the parent, skip
        if (child == parent) continue;

        // We only orient edges that are NOT bridges
        // Bridges will be printed in both directions anyway.
        // Check if (node, child) is a bridge
        bool bridge_edge = (is_bridge.count({min(node, child), max(node, child)}) > 0);

        if (!bridge_edge) {
            // If we haven't oriented this edge yet, do so
            // We'll orient from node to child if possible
            // We can also skip if child is already used_for_orientation
            // but we do want an orientation anyway.  
            // So let's just do it once: always direct node -> child in the DFS tree.
            // (If child's not visited, that means it's a DFS tree edge from node to child.)

            // We'll add a directed edge node->child if we haven't visited child yet
            // or even if it's visited, it's OK to have node->child as a back edge direction.
            directed_edges.push_back({node, child});
        }

        // Continue DFS if child not visited
        if (!used_for_orientation[child]) {
            orient_dfs(child, node);
        }
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int caseNo = 1;

    while (true) {
        cin >> n >> m;
        if (!cin || n == 0 && m == 0) {
            break;
        }

        // Clear data structures
        for(int i = 1; i <= n; i++) {
            graph[i].clear();
            visited[i] = false;
            used_for_orientation[i] = false;
        }
        edges.clear();
        bridges.clear();
        is_bridge.clear();
        
        timer = 1;

        // Read the edges
        // Intersections are from 1..n
        for (int i = 0; i < m; i++) {
            int u, v;
            cin >> u >> v;
            edges.push_back({u, v});
            // Build undirected adjacency
            graph[u].push_back(v);
            graph[v].push_back(u);
        }

        // 1) Find bridges
        // We assume the graph is connected as per the problem statement
        // But in case it's not, we run DFS from every unvisited node.
        for(int i = 1; i <= n; i++) {
            if(!visited[i]) {
                dfs_bridge(i, -1);
            }
        }

        // 2) Orient all non-bridge edges
        directed_edges.clear();
        // We run a DFS (or BFS) from node 1 (or from each component if needed)
        // to set orientation for non-bridge edges.
        for(int i = 1; i <= n; i++){
            if(!used_for_orientation[i]) {
                orient_dfs(i, -1);
            }
        }

        // Print output
        cout << caseNo++ << "\n\n";  // case number + blank line

        // For each original edge
        //   if it is a bridge, print "u v" and "v u"
        //   otherwise, we find the orientation we assigned (u->v in directed_edges).
        //   There's a simpler way: we can keep track of oriented edges in a map or set.
        //   We'll do it with a lookup:
        
        // Build a set of oriented edges for quick membership checking
        set<pair<int,int>> oriented;
        for (auto &e : directed_edges) {
            oriented.insert(e); 
        }
        
        for (auto &e : edges) {
            int u = e.first, v = e.second;
            // check if it's a bridge
            bool br = (is_bridge.count({min(u, v), max(u, v)}) > 0);
            if (br) {
                // print both directions
                cout << u << " " << v << "\n";
                cout << v << " " << u << "\n";
            } else {
                // We should have oriented u->v or v->u
                // check which direction was stored
                if (oriented.count({u, v})) {
                    cout << u << " " << v << "\n";
                } else {
                    // must be v->u
                    cout << v << " " << u << "\n";
                }
            }
        }

        cout << "#\n";
    }

    return 0;
}

Explanation of Key Parts

1. Bridge-finding DFS (dfs_bridge):

   • A[node] is set to the discovery time of node.
   • B[node] is initially the same as A[node].
   • For each unvisited child, we recurse and then update B[node] = min(B[node], B[child]).
   • If B[child] > A[node], then (node, child) is a bridge.

2. Orientation DFS (orient_dfs):

   • We traverse from a starting node (e.g., 1).
   • For every non-bridge edge (node, child), we orient it from node to child.
   • We store that in directed_edges.

3. Printing:

   • If an edge is a bridge, print it both ways: $$u\;v \quad \text{and} \quad v\;u.$$
   • Otherwise, print it in the single direction we found during the orientation DFS.

Because the graph is guaranteed to be connected in the problem statement, each test case’s orientation ensures that, for non-bridge edges, we give a single direction consistent with the DFS tree or back edges, while bridges necessarily appear in both directions.

This solves the stated requirement:

From each intersection, it is possible to drive to all the other intersections.

Any single orientation of a bridge would break strong connectivity, so we allow both directions in that case.