Solution for Message Routes

4_graph_algorithms/Message_Route.cpp

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+/*
+Problem: Message Route
+Category: Graph (Shortest Path - BFS)
+CSES Link: https://cses.fi/problemset/task/1667
+Difficulty: Medium
+Time Complexity: O(N + M)
+Space Complexity: O(N + M)
+
+Approach:
+Perform Breadth-First Search (BFS) from node 1 on the undirected, unweighted graph.
+BFS finds shortest distances (in edges) from the source to every reachable node.
+Track a `parent` array while exploring to reconstruct the shortest path from 1 to n.
+If node n is unreachable, print "IMPOSSIBLE". Otherwise print the number of nodes
+on the shortest route and the route itself.
+
+Key Insights:
+- BFS is ideal for shortest path in unweighted graphs because it explores level-by-level.
+- Use a large sentinel (INF) to mark unvisited nodes in the distance array.
+- Reconstruct path by following parent pointers from destination back to source.
+- Use non-recursive approach (BFS) to avoid recursion depth issues on large inputs.
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    // Read number of nodes (computers) and edges (connections)
+    int n, m;
+    cin >> n >> m;
+
+    // Build adjacency list (1-indexed)
+    vector<vector<int>> adj(n + 1);
+    for (int i = 0; i < m; ++i) {
+        int u, v;
+        cin >> u >> v;
+        // undirected edge
+        adj[u].push_back(v);
+        adj[v].push_back(u);
+    }
+
+    const int INF = 1e9;
+    vector<int> distance(n + 1, INF); // distance[i] = distance from node 1 to i
+    vector<int> parent(n + 1, -1);     // parent[i] = previous node on shortest path to i
+
+    // BFS initialization
+    queue<int> q;
+    distance[1] = 0;
+    q.push(1);
+
+    // Standard BFS loop
+    while (!q.empty()) {
+        int current = q.front();
+        q.pop();
+
+        for (int neighbor : adj[current]) {
+            // If neighbor not visited yet
+            if (distance[neighbor] == INF) {
+                distance[neighbor] = distance[current] + 1; // one more edge
+                parent[neighbor] = current;                 // record path
+                q.push(neighbor);
+            }
+        }
+    }
+
+    // If destination node n is unreachable, print IMPOSSIBLE
+    if (distance[n] == INF) {
+        cout << "IMPOSSIBLE\n";
+        return 0;
+    }
+
+    // Reconstruct path from n back to 1 using parent pointers
+    vector<int> path;
+    int node = n;
+    while (node != -1) {
+        path.push_back(node);
+        node = parent[node];
+    }
+    reverse(path.begin(), path.end());
+
+    // Print number of nodes on path and the path itself
+    cout << path.size() << '\n';
+    for (size_t i = 0; i < path.size(); ++i) {
+        if (i) cout << ' ';
+        cout << path[i];
+    }
+    cout << '\n';
+    return 0;
+}