Finding shortest path between any two nodes using Floyd Warshall Algorithm

Given a graph and two nodes u and v, the task is to print the shortest path between u and v using the Floyd Warshall algorithm.

Examples:

Input: u = 1, v = 3

Output: 1 -> 2 -> 3
Explanation:
Shortest path from 1 to 3 is through vertex 2 with total cost 3.
The first edge is 1 -> 2 with cost 2 and the second edge is 2 -> 3 with cost 1.

Input: u = 0, v = 2

Output: 0 -> 1 -> 2
Explanation:
Shortest path from 0 to 2 is through vertex 1 with total cost = 5

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

The main idea here is to use a matrix(2D array) that will keep track of the next node to point if the shortest path changes for any pair of nodes. Initially, the shortest path between any two nodes u and v is v (that is the direct edge from u -> v).
Initialising the Next array

If the path exists between two nodes then Next[u][v] = v
else we set Next[u][v] = -1
Modification in Floyd Warshall Algorithm

Inside the if condition of Floyd Warshall Algorithm we’ll add a statement Next[i][j] = Next[i][k]
(that means we found the shortest path between i, j through an intermediate node k).

This is how our if condition would look like
```
if(dis[i][j] > dis[i][k] + dis[k][j])
{
    dis[i][j] = dis[i][k] + dis[k][j];
    Next[i][j] = Next[i][k];    
}
```
For constructing path using these nodes we’ll simply start looping through the node u while updating its value to next[u][v] until we reach node v.
```
path = [u]
while u != v:
    u = Next[u][v]
    path.append(u)
```

Below is the implementation of the above approach.

C++

// C++ program to find the shortest  
// path between any two nodes using  
// Floyd Warshall Algorithm.  
#include <bits/stdc++.h>  
using namespace std;  
  
#define MAXN 100  
// Infinite value for array  
const int INF = 1e7;  
  
int dis[MAXN][MAXN];  
int Next[MAXN][MAXN];  
  
// Initializing the distance and  
// Next array  
void initialise(int V,  
                vector<vector<int> >& graph)  
{  
    for (int i = 0; i < V; i++) {  
        for (int j = 0; j < V; j++) {  
            dis[i][j] = graph[i][j];  
  
            // No edge between node  
            // i and j  
            if (graph[i][j] == INF)  
                Next[i][j] = -1;  
            else
                Next[i][j] = j;  
        }  
    }  
}  
  
// Function construct the shotest  
// path between u and v  
vector<int> constructPath(int u,  
                        int v)  
{  
    // If there's no path between  
    // node u and v, simply return  
    // an empty array  
    if (Next[u][v] == -1)  
        return {};  
  
    // Storing the path in a vector  
    vector<int> path = { u };  
    while (u != v) {  
        u = Next[u][v];  
        path.push_back(u);  
    }  
    return path;  
}  
  
// Standard Floyd Warshall Algorithm  
// with little modification Now if we find  
// that dis[i][j] > dis[i][k] + dis[k][j]  
// then we modify next[i][j] = next[i][k]  
void floydWarshall(int V)  
{  
    for (int k = 0; k < V; k++) {  
        for (int i = 0; i < V; i++) {  
            for (int j = 0; j < V; j++) {  
  
                // We cannot travel through  
                // edge that doesn't exist  
                if (dis[i][k] == INF  
                    || dis[k][j] == INF)  
                    continue;  
  
                if (dis[i][j] > dis[i][k]  
                                    + dis[k][j]) {  
                    dis[i][j] = dis[i][k]  
                                + dis[k][j];  
                    Next[i][j] = Next[i][k];  
                }  
            }  
        }  
    }  
}  
  
// Print the shortest path  
void printPath(vector<int>& path)  
{  
    int n = path.size();  
    for (int i = 0; i < n - 1; i++)  
        cout << path[i] << " -> ";  
    cout << path[n - 1] << endl;  
}  
  
// Driver code  
int main()  
{  
  
    int V = 4;  
    vector<vector<int> > graph  
        = { { 0, 3, INF, 7 },  
            { 8, 0, 2, INF },  
            { 5, INF, 0, 1 },  
            { 2, INF, INF, 0 } };  
  
    // Function to initialise the  
    // distance and Next array  
    initialise(V, graph);  
  
    // Calling Floyd Warshall Algorithm,  
    // this will update the shortest  
    // distance as well as Next array  
    floydWarshall(V);  
    vector<int> path;  
  
    // Path from node 1 to 3  
    cout << "Shortest path from 1 to 3: ";  
    path = constructPath(1, 3);  
    printPath(path);  
  
    // Path from node 0 to 2  
    cout << "Shortest path from 0 to 2: ";  
    path = constructPath(0, 2);  
    printPath(path);  
  
    // path from node 3 to 2  
    cout << "Shortest path from 3 to 2: ";  
    path = constructPath(3, 2);  
    printPath(path);  
  
    return 0;  
}  

Java

// Java program to find the shortest  
// path between any two nodes using  
// Floyd Warshall Algorithm.  
import java.util.*;  
  
class GFG{  
  
static final int MAXN = 100;  
  
// Infinite value for array  
static int INF = (int) 1e7;  
  
static int [][]dis = new int[MAXN][MAXN];  
static int [][]Next = new int[MAXN][MAXN];  
  
// Initializing the distance and  
// Next array  
static void initialise(int V,  
                    int [][] graph)  
{  
    for(int i = 0; i < V; i++)  
    {  
    for(int j = 0; j < V; j++)  
    {  
        dis[i][j] = graph[i][j];  
              
        // No edge between node  
        // i and j  
        if (graph[i][j] == INF)  
            Next[i][j] = -1;  
        else
            Next[i][j] = j;  
    }  
    }  
}  
  
// Function conthe shotest  
// path between u and v  
static Vector<Integer> constructPath(int u,  
                                    int v)  
{  
  
    // If there's no path between  
    // node u and v, simply return  
    // an empty array  
    if (Next[u][v] == -1)  
        return null;  
  
    // Storing the path in a vector  
    Vector<Integer> path = new Vector<Integer>();  
    path.add(u);  
      
    while (u != v)  
    {  
        u = Next[u][v];  
        path.add(u);  
    }  
    return path;  
}  
  
// Standard Floyd Warshall Algorithm  
// with little modification Now if we find  
// that dis[i][j] > dis[i][k] + dis[k][j]  
// then we modify next[i][j] = next[i][k]  
static void floydWarshall(int V)  
{  
    for(int k = 0; k < V; k++)  
    {  
    for(int i = 0; i < V; i++)  
    {  
        for(int j = 0; j < V; j++)  
        {  
              
            // We cannot travel through  
            // edge that doesn't exist  
            if (dis[i][k] == INF ||  
                dis[k][j] == INF)  
                continue;  
                  
            if (dis[i][j] > dis[i][k] +  
                            dis[k][j])  
            {  
                dis[i][j] = dis[i][k] +  
                            dis[k][j];  
                Next[i][j] = Next[i][k];  
            }  
        }  
    }  
    }  
}  
  
// Print the shortest path  
static void printPath(Vector<Integer> path)  
{  
    int n = path.size();  
    for(int i = 0; i < n - 1; i++)  
    System.out.print(path.get(i) + " -> ");  
    System.out.print(path.get(n - 1) + "\n");  
}  
  
// Driver code  
public static void main(String[] args)  
{  
    int V = 4;  
    int [][] graph = { { 0, 3, INF, 7 },  
                    { 8, 0, 2, INF },  
                    { 5, INF, 0, 1 },  
                    { 2, INF, INF, 0 } };  
  
    // Function to initialise the  
    // distance and Next array  
    initialise(V, graph);  
  
    // Calling Floyd Warshall Algorithm,  
    // this will update the shortest  
    // distance as well as Next array  
    floydWarshall(V);  
    Vector<Integer> path;  
  
    // Path from node 1 to 3  
    System.out.print("Shortest path from 1 to 3: ");  
    path = constructPath(1, 3);  
    printPath(path);  
  
    // Path from node 0 to 2  
    System.out.print("Shortest path from 0 to 2: ");  
    path = constructPath(0, 2);  
    printPath(path);  
  
    // Path from node 3 to 2  
    System.out.print("Shortest path from 3 to 2: ");  
    path = constructPath(3, 2);  
    printPath(path);  
}  
}  
  
// This code is contributed by Amit Katiyar  

C#

// C# program to find the shortest 
// path between any two nodes using 
// Floyd Warshall Algorithm. 
using System; 
using System.Collections.Generic; 
  
class GFG{ 
  
static readonly int MAXN = 100; 
  
// Infinite value for array 
static int INF = (int)1e7; 
  
static int [,]dis = new int[MAXN, MAXN]; 
static int [,]Next = new int[MAXN, MAXN]; 
  
// Initializing the distance and 
// Next array 
static void initialise(int V, 
                       int [,] graph) 
{ 
    for(int i = 0; i < V; i++) 
    { 
        for(int j = 0; j < V; j++)  
        { 
            dis[i, j] = graph[i, j]; 
                  
            // No edge between node 
            // i and j 
            if (graph[i, j] == INF) 
                Next[i, j] = -1; 
            else
                Next[i, j] = j; 
        } 
    } 
} 
  
// Function conthe shotest 
// path between u and v 
static List<int> constructPath(int u, int v) 
{ 
      
    // If there's no path between 
    // node u and v, simply return 
    // an empty array 
    if (Next[u, v] == -1) 
        return null; 
  
    // Storing the path in a vector 
    List<int> path = new List<int>(); 
    path.Add(u); 
      
    while (u != v)  
    { 
        u = Next[u, v]; 
        path.Add(u); 
    } 
    return path; 
} 
  
// Standard Floyd Warshall Algorithm 
// with little modification Now if we find 
// that dis[i,j] > dis[i,k] + dis[k,j] 
// then we modify next[i,j] = next[i,k] 
static void floydWarshall(int V) 
{ 
    for(int k = 0; k < V; k++) 
    { 
        for(int i = 0; i < V; i++) 
        { 
            for(int j = 0; j < V; j++) 
            { 
                  
                // We cannot travel through  
                // edge that doesn't exist  
                if (dis[i, k] == INF ||   
                    dis[k, j] == INF)  
                    continue;  
                     
                if (dis[i, j] > dis[i, k] +  
                                dis[k, j])  
                {  
                    dis[i, j] = dis[i, k] +   
                                dis[k, j];  
                    Next[i, j] = Next[i, k];  
                }  
            } 
        } 
    } 
} 
  
// Print the shortest path 
static void printPath(List<int> path) 
{ 
    int n = path.Count; 
      
    for(int i = 0; i < n - 1; i++)  
        Console.Write(path[i] + " -> "); 
         
    Console.Write(path[n - 1] + "\n"); 
} 
  
// Driver code 
public static void Main(String[] args) 
{ 
    int V = 4; 
    int [,] graph = { { 0, 3, INF, 7 }, 
                      { 8, 0, 2, INF }, 
                      { 5, INF, 0, 1 }, 
                      { 2, INF, INF, 0 } }; 
  
    // Function to initialise the 
    // distance and Next array 
    initialise(V, graph); 
  
    // Calling Floyd Warshall Algorithm, 
    // this will update the shortest 
    // distance as well as Next array 
    floydWarshall(V); 
    List<int> path; 
  
    // Path from node 1 to 3 
    Console.Write("Shortest path from 1 to 3: "); 
    path = constructPath(1, 3); 
    printPath(path); 
  
    // Path from node 0 to 2 
    Console.Write("Shortest path from 0 to 2: "); 
    path = constructPath(0, 2); 
    printPath(path); 
  
    // Path from node 3 to 2 
    Console.Write("Shortest path from 3 to 2: "); 
    path = constructPath(3, 2); 
    printPath(path); 
} 
} 
  
// This code is contributed by Amit Katiyar 

Output:

Shortest path from 1 to 3: 1 -> 2 -> 3
Shortest path from 0 to 2: 0 -> 1 -> 2
Shortest path from 3 to 2: 3 -> 0 -> 1 -> 2

Complexity Analysis:

The time complexity for Floyd Warshall Algorithm is O(V³)
For finding shortest path time complexity is O(V) per query.

Note: It would be efficient to use the Floyd Warshall Algorithm when your graph contains a couple of hundred vertices and you need to answer multiple queries related to the shortest path.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

C#

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