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+/*
+Problem: Shortest Routes I
+Category: Graph Algorithms (Single Source Shortest Path)
+Difficulty: Medium
+Time Complexity: O((n + m) * log n)
+Space Complexity: O(n + m)
+
+Approach:
+This problem asks for the shortest distance from a source city (1) to every other city in a directed, weighted graph.
+Since all edge weights are positive, Dijkstra’s Algorithm is the optimal approach.
+
+Steps:
+1. Represent the graph using an adjacency list where each node stores pairs (neighbor, weight).
+2. Initialize all distances as infinity (INF = 1e18) and set the source city’s distance (city 1) to 0.
+3. Use a min-heap (priority queue) to always pick the node with the smallest current distance.
+4. For each node popped from the queue:
+   - If the current distance is greater than the stored distance, skip (outdated entry).
+   - For all adjacent nodes, relax the edge: 
+       if dist[u] + w < dist[v], update dist[v] and push (dist[v], v) into the priority queue.
+5. After processing all nodes, print the distances from city 1 to every city in order.
+
+Key Insights:
+- Dijkstra’s algorithm is efficient for graphs with **non-negative weights**, unlike Bellman-Ford.
+- Priority queue ensures that each edge relaxation happens optimally, minimizing redundant work.
+- Since the problem guarantees that all nodes are reachable from city 1, there’s no need to handle disconnected components.
+- Using `long long` (64-bit integers) prevents overflow, as path lengths can exceed 1e9.
+- Fast I/O (`ios::sync_with_stdio(false); cin.tie(nullptr);`) is essential due to large input size.
+- Memory-efficient representation with adjacency list handles up to 2×10^5 edges comfortably.
+
+Optimization Tricks:
+- Avoid reprocessing nodes by checking `if (d > dist[u]) continue;`.
+- Store distances in a vector instead of an unordered_map for faster access (since node indices are contiguous).
+- Use `greater<pair<int, int>>` with priority_queue for min-heap behavior.
+
+Edge Cases:
+- Multiple edges between same nodes → Dijkstra naturally handles it (shorter path replaces longer one).
+- Single city (n = 1) → Output is just 0.
+- Large input graphs → Must ensure O((n + m) log n) implementation to stay within 1s time limit.
+
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+#define int long long
+const int INF = 1e18;  // Large value representing infinity
+
+int32_t main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, m;
+    cin >> n >> m;
+
+    vector<vector<pair<int, int>>> adj(n + 1); // {neighbor, weight}
+
+    for (int i = 0; i < m; i++) {
+        int a, b, c;
+        cin >> a >> b >> c;
+        adj[a].push_back({b, c});
+    }
+
+    vector<int> dist(n + 1, INF);
+    dist[1] = 0; // Distance to source is 0
+
+    // Min-heap: {distance, node}
+    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
+    pq.push({0, 1});
+
+    while (!pq.empty()) {
+        auto [d, u] = pq.top();
+        pq.pop();
+
+        if (d > dist[u]) continue; // Skip outdated entry
+
+        for (auto [v, w] : adj[u]) {
+            if (dist[u] + w < dist[v]) {
+                dist[v] = dist[u] + w;
+                pq.push({dist[v], v});
+            }
+        }
+    }
+
+    for (int i = 1; i <= n; i++) {
+        cout << dist[i] << " ";
+    }
+    cout << "\n";
+
+    return 0;
+}