Finding shortest path between any two nodes using Floyd Warshall Algorithm

Difficulty Level : Medium
Last Updated : 13 Jan, 2021

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

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

Python3

# Python3 program to find the shortest
# path between any two nodes using
# Floyd Warshall Algorithm.
 
# Initializing the distance and
# Next array
def initialise(V):
    global dis, Next
 
    for i in range(V):
        for j in range(V):
            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
def constructPath(u, v):
    global graph, Next
     
    # 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
    path = [u]
    while (u != v):
        u = Next[u][v]
        path.append(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]
def floydWarshall(V):
    global dist, Next
    for k in range(V):
        for i in range(V):
            for j in range(V):
                 
                # We cannot travel through
                # edge that doesn't exist
                if (dis[i][k] == INF or 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]
 
# Prthe shortest path
def printPath(path):
    n = len(path)
    for i in range(n - 1):
        print(path[i], end=" -> ")
    print (path[n - 1])
 
# Driver code
if __name__ == '__main__':
    MAXM,INF = 100,10**7
    dis = [[-1 for i in range(MAXM)] for i in range(MAXM)]
    Next = [[-1 for i in range(MAXM)] for i in range(MAXM)]
 
    V = 4
    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)
 
    # Calling Floyd Warshall Algorithm,
    # this will update the shortest
    # distance as well as Next array
    floydWarshall(V)
    path = []
 
    # Path from node 1 to 3
    print("Shortest path from 1 to 3: ", end = "")
    path = constructPath(1, 3)
    printPath(path)
 
    # Path from node 0 to 2
    print("Shortest path from 0 to 2: ", end = "")
    path = constructPath(0, 2)
    printPath(path)
 
    # Path from node 3 to 2
    print("Shortest path from 3 to 2: ", end = "")
    path = constructPath(3, 2)
    printPath(path)
 
    # This code is contributed by mohit kumar 29

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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