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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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Python3

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# 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

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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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