- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
All-Pairs Shortest Paths
The all pair shortest path algorithm is also known as Floyd-Warshall algorithm is used to find all pair shortest path problem from a given weighted graph. As a result of this algorithm, it will generate a matrix, which will represent the minimum distance from any node to all other nodes in the graph.
At first the output matrix is same as given cost matrix of the graph. After that the output matrix will be updated with all vertices k as the intermediate vertex.
The time complexity of this algorithm is O(V3), here V is the number of vertices in the graph.
Input − The cost matrix of the graph.
0 3 6 ∞ ∞ ∞ ∞ 3 0 2 1 ∞ ∞ ∞ 6 2 0 1 4 2 ∞ ∞ 1 1 0 2 ∞ 4 ∞ ∞ 4 2 0 2 1 ∞ ∞ 2 ∞ 2 0 1 ∞ ∞ ∞ 4 1 1 0
Output − Matrix of all pair shortest path.
0 3 4 5 6 7 7 3 0 2 1 3 4 4 4 2 0 1 3 2 3 5 1 1 0 2 3 3 6 3 3 2 0 2 1 7 4 2 3 2 0 1 7 4 3 3 1 1 0
Algorithm
floydWarshal(cost)
Input − The cost matrix of given Graph.
Output − Matrix to for shortest path between any vertex to any vertex.
Begin for k := 0 to n, do for i := 0 to n, do for j := 0 to n, do if cost[i,k] + cost[k,j] < cost[i,j], then cost[i,j] := cost[i,k] + cost[k,j] done done done display the current cost matrix End
Example
#include<iostream> #include<iomanip> #define NODE 7 #define INF 999 using namespace std; //Cost matrix of the graph int costMat[NODE][NODE] = { {0, 3, 6, INF, INF, INF, INF}, {3, 0, 2, 1, INF, INF, INF}, {6, 2, 0, 1, 4, 2, INF}, {INF, 1, 1, 0, 2, INF, 4}, {INF, INF, 4, 2, 0, 2, 1}, {INF, INF, 2, INF, 2, 0, 1}, {INF, INF, INF, 4, 1, 1, 0} }; void floydWarshal(){ int cost[NODE][NODE]; //defind to store shortest distance from any node to any node for(int i = 0; i<NODE; i++) for(int j = 0; j<NODE; j++) cost[i][j] = costMat[i][j]; //copy costMatrix to new matrix for(int k = 0; k<NODE; k++){ for(int i = 0; i<NODE; i++) for(int j = 0; j<NODE; j++) if(cost[i][k]+cost[k][j] < cost[i][j]) cost[i][j] = cost[i][k]+cost[k][j]; } cout << "The matrix:" << endl; for(int i = 0; i<NODE; i++){ for(int j = 0; j<NODE; j++) cout << setw(3) << cost[i][j]; cout << endl; } } int main(){ floydWarshal(); }
Output
The matrix: 0 3 5 4 6 7 7 3 0 2 1 3 4 4 5 2 0 1 3 2 3 4 1 1 0 2 3 3 6 3 3 2 0 2 1 7 4 2 3 2 0 1 7 4 3 3 1 1 0