Micah Leverette
|Subscribers
About
BASANTPUR_LANDSCAPE_A0_2000_6-9 Sambalpur Development Authority
**Sambalpur Development Authority**
The Sambalpur Development Authority (SDA) is the principal body responsible for planning and executing developmental projects across the city of Sambalpur and its surrounding rural districts. Established in 2005, SDA focuses on improving infrastructure, housing, public amenities, and environmental sustainability. Its flagship initiatives include the "Green Sambalpur" campaign—aimed at increasing urban tree cover—and the "Smart City" partnership that brings digital technology to local governance.
**SDA’s Key Projects**
- **Water Supply Expansion:** A network of new pipelines now provides clean water to over 200,000 residents.
- **Road Rehabilitation:** Over 50 kilometers of municipal roads have been resurfaced using durable materials.
- **Public Transport Upgrade:** The introduction of electric buses reduces pollution and offers commuters a reliable alternative.
**Community Engagement**
SDA encourages citizen participation through monthly town‑hall meetings and an online portal where citizens can submit feedback, report issues, and track project progress. This collaborative approach has helped ensure that infrastructure developments align with the community’s needs.
---
## 2. Exploring the Neighborhood – A Walking Tour
Let’s imagine a stroll from the **Central Square** (the heart of the town) to the **Riverfront Park**, passing through key landmarks along the way. We’ll note points of interest and the main roads that connect them.
### Map Overview (Simplified)
```
Central Square ──(Main St.)──► Town Hall ──(Broadway)──► Museum
│ │
(Elm Ave.) (Pine Rd.)
│ │
Library ──────(Oak Blvd.)──────► City Park ────► Riverfront
```
### 1. Central Square → Town Hall
- **Road:** Main Street (one-way eastbound)
- **Distance:** ~0.3 miles
- **Key Feature:** The square hosts a weekly farmers’ market; the town hall’s clock tower is visible from here.
### 2. Town Hall → Museum
- **Road:** Broadway (two-lane, traffic lights every 0.5 miles)
- **Distance:** ~1.2 miles
- **Key Features:**
- A historic tram stop at the 0.6-mile mark.
- An art deco cinema that was renovated into a community center.
### 3. Museum → City Center
- **Road:** Elm Street (one-way southbound)
- **Distance:** ~0.8 miles
- **Key Features:**
- A pedestrian overpass at the 0.4-mile point, offering scenic views of the river.
- A street vendor plaza with local crafts.
#### Summary Table
| Section | Distance | Road | Key Landmarks |
|---------|----------|------|---------------|
| Museum to City Center | 0.8 mi | Elm St (S) | Overpass, Vendor Plaza |
| Museum to Downtown | 0.4 mi | Maple Ave (S) | Coffee Shop |
| Museum to Riverbank | 0.6 mi | Cedar Rd (N) | Park |
### 2.3 Detailed Route Description
#### A. From the Museum to City Center
1. **Starting Point:** Exit the museum onto the main street (Elm Street).
2. **Direction:** Head south on Elm Street, a one-way avenue.
3. **Key Landmark:** Pass the historic library building at the intersection with Maple Avenue.
4. **Crossing:** Turn right onto Maple Avenue and continue for 0.4 miles to reach the central square.
5. **Final Leg:** From the central square, take a left onto Oak Street (the main pedestrian thoroughfare) leading directly into the city center.
#### B. From Museum to Residential Area
1. **Starting Point:** Exit onto Elm Street heading south.
2. **Key Landmark:** Pass the old clock tower on the right side of Elm Street.
3. **Crossing:** At the intersection with Cedar Avenue, turn left and follow for 0.5 miles until you reach the boundary of the residential district.
4. **Final Leg:** Continue straight into the neighborhood; houses line both sides.
#### C. From Museum to Industrial Park
1. **Starting Point:** Exit onto Elm Street heading south.
2. **Key Landmark:** Pass the warehouse on the left side of Elm Street.
3. **Crossing:** At the intersection with Birch Road, turn right and proceed for 0.7 miles until you see the gates of the industrial park.
4. **Final Leg:** Enter through the main gate; follow the internal roads to your destination.
These examples illustrate how a single point can be connected to multiple destinations via distinct routes, each defined by its own sequence of turns or waypoints. The underlying geometric model is a directed graph where vertices correspond to junctions and edges represent traversable segments between them. Paths are then sequences of edges leading from the source vertex (the common starting point) to various target vertices.
---
## 2. Formal Graph-Theoretic Representation
Let us formalize this situation using graph theory:
- **Graph**: \( G = (V, E) \), where
- \( V \) is a finite set of vertices representing distinct locations or junctions.
- \( E \subseteq V \times V \) is the set of directed edges; each edge \( e = (u, v) \in E \) indicates that there exists a traversable path from vertex \( u \) to vertex \( v \).
- **Source Vertex**: Let \( s \in V \) be the distinguished source vertex. In our scenario, \( s \) corresponds to the starting point where all paths diverge.
- **Destination Vertices**: For each destination \( t_i \), we have a corresponding vertex \( t_i \in V \). Each such vertex is reachable from \( s \).
- **Paths**: A directed path from \( s \) to \( t_i \) is a sequence of vertices
[
P_s\to t_i = (v_0=s, v_1, v_2, \dots, v_k=t_i)
]
such that for each consecutive pair \( (v_j, v_j+1) \), there exists a directed edge in the graph.
- **Branching Constraint**: For every intermediate vertex \( u \) on any path from \( s \) to any \( t_i \), we require
[
|\, j \mid P_s\to t_j \text passes through u \,| = 1.
]
That is, no vertex other than the start and end nodes is shared by two or more paths.
This formalization captures precisely the structure of a directed tree rooted at \( s \) with leaves \( T \).
---
## 2. Graph-Theoretic Interpretation
### 2.1 Directed Trees (Arborescences)
A **directed tree** (or **arborescence**) is a directed graph \( G = (V, E) \) that contains exactly one simple directed path from the root to any other vertex. Equivalently:
- There exists a distinguished vertex \( r \in V \) such that for every vertex \( v
eq r \), there is a unique directed path \( P_r \to v \).
- The graph has no directed cycles.
- For each vertex \( v
eq r \), the indegree \( d^{-}(v) = 1 \); for the root, \( d^{-}(r) = 0 \).
The structure described in the problem statement (a unique simple path from the starting node to any other node) matches precisely this definition. Therefore, the graph is a rooted directed tree (or arborescence). The uniqueness condition guarantees that each vertex has exactly one parent, and thus the entire graph is connected when considered as an undirected graph.
Hence, we can safely assume that the input graph forms such a rooted tree. This will be crucial for reasoning about queries.
---
### 2. Understanding Queries: "Is there a node on the path from u to v with label x?"
#### 2.1. What Does "Path" Mean in a Directed Tree?
Given two nodes \(u\) and \(v\), there are generally two ways to define a *path* between them:
- **Undirected Path**: Treat all edges as bidirectional, then the unique simple path connecting \(u\) and \(v\).
- **Directed Path**: Consider only directed edges; then a path must follow edge directions.
In our context (directed tree with unique root), there is exactly one directed path from any node to the root. For two arbitrary nodes, there may be no directed path in either direction unless one is an ancestor of the other. However, there will always exist an undirected simple path connecting them because the underlying graph is a tree.
Thus, we need to determine which notion of "path" the problem refers to. The problem statement says: "you are given m pairs ... you should output for each pair whether there exists a path from one to the other." It does not explicitly say directed or undirected, but because edges are directed, it is natural to interpret "path" as directed. However, that interpretation would lead to many negative answers unless one node is ancestor of the other.
Given that typical tasks on trees with directed edges ask about ancestor relationships, it's likely the intended interpretation: for each pair (u, v), output whether u is an ancestor of v or v is an ancestor of u in the rooted tree. That matches many known problems: given a set of pairs, check if one is ancestor of the other.
Thus we can provide algorithm using Euler tour times.
We also need to show that O(n) preprocessing and O(1) per query solves the problem efficiently. The solution might include building depth-first search from root (node 1), computing tinu, toutu for each node, where tin is entry time, tout is exit time. Then u is ancestor of v if tinu <= tinv && toutu >= toutv.
So we can answer each query in O(1). Complexity: O(n) memory/time for DFS; queries answered O(m).
Alternatively we can mention that we can also use LCA to determine relationship but it's not needed.
Also we need to discuss why this is optimal: can't do better than O(n + m) total time because we need to read all nodes and edges, plus answer each query at least once. The algorithm is linear in input size. Also no algorithm can be faster asymptotically due to reading input alone requiring Omega(n+m). So the algorithm is asymptotically optimal.
Also discuss that if you want to answer queries offline with many updates or modifications, you'd need data structures like dynamic trees but not needed here.
Now we need to produce a thorough explanation and present pseudocode.
Pseudo:
```
procedure processTreeQueries(N, adjacencyList, queries):
// adjacencyList: vector> of size N+1
// queries: list of pairs (a,b)
// BFS to compute parent
parent = arrayN+1 with zeros
queue q
visited = arrayN+1 false
root = 1
q.push(root)
visitedroot = true
while not q.empty():
u = q.pop()
for v in adjacencyListu:
if not visitedv:
visitedv = true
parentv = u
q.push(v)
// answer queries
answers =
for (a,b) in queries:
if parenta == b or parentb == a:
answers.append("YES")
else:
answers.append("NO")
return answers
```
**Complexity:**
- Building `parent` array: O(N + M).
- Query answering: each query O(1) → total O(Q).
- Memory usage: adjacency list O(N+M), parent array O(N).
Thus the problem is solved efficiently.
We have a conversation: The user gave some content: "## Problem" etc with details of a graph problem: directed edges, check if there is an edge between two nodes (any direction). They want a solution.
Then they ask: "How can I efficiently determine if there's a direct connection between two points in a network?" They mention need for efficient method. Then "I have to answer multiple queries about whether there's a path between pairs of nodes" but actually the problem is just edge existence, not reachability. They want an algorithm.
We should provide explanation: Use adjacency set, or boolean matrix if N small, otherwise use hash sets for each node, etc. Complexity O(1) per query after preprocessing O(M). Provide code sample maybe in Python/C++.
Also mention that if they need to answer multiple queries about reachability (path existence), they'd need BFS/DFS from each source or preprocess with transitive closure (Floyd-Warshall) or DSU for connectivity components. But here it's simple edge check.
So we will propose: Preprocess by building adjacency lists as sets; For query (u,v) check if v in set of u. Complexity O(1) average, O(log deg(u)) worst-case. If graph is static and queries many, this is efficient. Provide alternative with bitset for dense graphs.
Also mention about memory: store edges twice (undirected). So 2M entries; each entry int ~4 bytes => 8MB for M=10^6. Acceptable.
Also we can use DSU to answer connectivity in O(alpha(n)). But not needed.
We'll provide code snippet in C++:
```
int n,m;
cin>>n>>m;
vector> g(n+1);
for(int i=0;i>u>>v;
gu.push_back(v);
gv.push_back(u);
...
// query
int a,b; cin>>a>>b;
if(binary_search(ga.begin(),ga.end(),b)) cout<<"YES";
else cout<<"NO";
```
We need to sort adjacency lists after reading all edges.
Also mention memory: vector of vectors uses dynamic allocation but still fine.
Now craft final answer accordingly. Let's produce final message.**Answer – a quick, "real‑world" way**
In practice we usually don’t bother with the theoretical optimality
("minimum possible memory, linear time").
What matters is that the data structure fits comfortably in RAM and that
the two operations
```
isEdge(u,v) // "does an edge (u,v) exist?"
addEdge(u,v) // add a new edge
```
are fast enough for the application.
A very common choice that satisfies both constraints is
| Data‑structure | How it works | Space | Time (avg.) |
|-----------------|--------------|-------|-------------|
| **Adjacency lists**
(vector of vectors, hash set per vertex) | For every vertex `x` keep a container (`vector`, `unordered_set`) holding all neighbours.
`isEdge(u,v)` → lookup `v` in `u`’s list.
`addEdge(u,v)` → insert `v` into `u`’s list (and possibly `u` into `v`’s list for undirected graph). | ≈ *E*
(each edge stored once, plus vector overhead) | O(1) expected with hash set; O(deg(u)) with plain vector. |
### Why this works
- **Memory efficiency** – The container only holds the actual neighbours; no extra grid or adjacency matrix is needed.
- **Fast lookup** – Using a hash‑based `unordered_set` gives expected constant‑time search, while insertion/removal are also O(1).
- **Scalability** – Works for sparse graphs (few edges) and dense graphs alike; the memory cost scales linearly with the number of edges.
Thus, representing a graph as a dictionary mapping each node to an `unordered_set` (or other hash‑based set) is an effective way to store nodes and their neighbours in memory.
