Sheldon And Pair Of Cities
Logical Breakdown
- [ ] Subproblem Identification:
- [ ] Optimal Substructure:
- [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
- [ ] Optimization: (Matrix Exponentiation if 'A' is huge)
Mathematical Rigor
State Definition
Let \(dp[i]\) be the state for the \(i\)-th subproblem.
Recurrence Relation
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.graphs;
import java.util.*;
/*
* ============================================================================================
* PROBLEM: Sheldon and Pair of Cities (All-Pairs Shortest Path)
* ============================================================================================
*
* ---------------------
* 1. Problem Description
* ---------------------
* Sheldon needs to find the shortest distance between multiple pairs of cities (G[i], H[i]).
* The country has A cities and B bidirectional roads with distance F[i].
* If cities are unreachable, return -1.
*
* ---------------------
* 2. Logical Breakdown & Mathematical Formulation
* ---------------------
* Since we have many queries (C) and need distances between arbitrary pairs, we use
* the Floyd-Warshall Algorithm to precompute the shortest path between every city pair.
*
* Step 1: Adjacency Matrix Initialization
* - Create a 2D array dist[A+1][A+1] initialized to a large value (INF).
* - For all i, dist[i][i] = 0.
* - For each road (u, v) with weight w:
* - Because there can be multiple roads between the same two cities, only keep
* the minimum weight: dist[u][v] = dist[v][u] = min(dist[u][v], w).
*
* Step 2: Floyd-Warshall DP
* - Use a triple nested loop to update distances via an intermediate vertex 'k':
* $\(dist[i][j] = \min(dist[i][j], dist[i][k] + dist[k][j])\)$
* - This effectively considers every city as a potential "shortcut" between all other pairs.
*
* Step 3: Query Handling
* - For each query (G[i], H[i]), if dist[G[i]][H[i]] is still INF, return -1.
* - Otherwise, return the precomputed distance.
*
* ---------------------
* 3. Complexity Analysis
* ---------------------
* Time Complexity:
* - Preprocessing: O(A^3) - The triple nested loop.
* - Queries: O(C) - Constant time lookup per query.
* Total: O(A^3 + C).
*
* Space Complexity:
* O(A^2) to store the all-pairs distance matrix.
* ============================================================================================
/
public class SheldonAndPairOfCities {
public int solve(int A) {
// TODO: Implement solution
return 0;
}
public int[] solve(int a, int b, int i, int[] d, int[] e, int[] f, int[] g, int[] h) {
return new int[0];
}
}