Floyd Warshall Algorithm | DP-16

Difficulty Level : Medium
Last Updated : 29 Jan, 2021

The Floyd Warshall Algorithm is for solving the All Pairs Shortest Path problem. The problem is to find shortest distances between every pair of vertices in a given edge weighted directed Graph.
Example:

Input:
       graph[][] = { {0,   5,  INF, 10},
                    {INF,  0,  3,  INF},
                    {INF, INF, 0,   1},
                    {INF, INF, INF, 0} }
which represents the following graph
             10
       (0)------->(3)
        |         /|\
      5 |          |
        |          | 1
       \|/         |
       (1)------->(2)
            3       
Note that the value of graph[i][j] is 0 if i is equal to j 
And graph[i][j] is INF (infinite) if there is no edge from vertex i to j.

Output:
Shortest distance matrix
      0      5      8      9
    INF      0      3      4
    INF    INF      0      1
    INF    INF    INF      0

Floyd Warshall Algorithm
We initialize the solution matrix same as the input graph matrix as a first step. Then we update the solution matrix by considering all vertices as an intermediate vertex. The idea is to one by one pick all vertices and updates all shortest paths which include the picked vertex as an intermediate vertex in the shortest path. When we pick vertex number k as an intermediate vertex, we already have considered vertices {0, 1, 2, .. k-1} as intermediate vertices. For every pair (i, j) of the source and destination vertices respectively, there are two possible cases.
1) k is not an intermediate vertex in shortest path from i to j. We keep the value of dist[i][j] as it is.
2) k is an intermediate vertex in shortest path from i to j. We update the value of dist[i][j] as dist[i][k] + dist[k][j] if dist[i][j] > dist[i][k] + dist[k][j]
The following figure shows the above optimal substructure property in the all-pairs shortest path problem.

Following is implementations of the Floyd Warshall algorithm.

C++

// C++ Program for Floyd Warshall Algorithm 
#include <bits/stdc++.h>
using namespace std;
 
// Number of vertices in the graph 
#define V 4 
 
/* Define Infinite as a large enough
value.This value will be used for 
vertices not connected to each other */
#define INF 99999 
 
// A function to print the solution matrix 
void printSolution(int dist[][V]); 
 
// Solves the all-pairs shortest path 
// problem using Floyd Warshall algorithm 
void floydWarshall (int graph[][V]) 
{ 
    /* dist[][] will be the output matrix 
    that will finally have the shortest 
    distances between every pair of vertices */
    int dist[V][V], i, j, k; 
 
    /* Initialize the solution matrix same 
    as input graph matrix. Or we can say 
    the initial values of shortest distances
    are based on shortest paths considering 
    no intermediate vertex. */
    for (i = 0; i < V; i++) 
        for (j = 0; j < V; j++) 
            dist[i][j] = graph[i][j]; 
 
    /* Add all vertices one by one to 
    the set of intermediate vertices. 
    ---> Before start of an iteration, 
    we have shortest distances between all 
    pairs of vertices such that the 
    shortest distances consider only the 
    vertices in set {0, 1, 2, .. k-1} as
    intermediate vertices. 
    ----> After the end of an iteration, 
    vertex no. k is added to the set of 
    intermediate vertices and the set becomes {0, 1, 2, .. k} */
    for (k = 0; k < V; k++) 
    { 
        // Pick all vertices as source one by one 
        for (i = 0; i < V; i++) 
        { 
            // Pick all vertices as destination for the 
            // above picked source 
            for (j = 0; j < V; j++) 
            { 
                // If vertex k is on the shortest path from 
                // i to j, then update the value of dist[i][j] 
                if (dist[i][k] + dist[k][j] < dist[i][j]) 
                    dist[i][j] = dist[i][k] + dist[k][j]; 
            } 
        } 
    } 
 
    // Print the shortest distance matrix 
    printSolution(dist); 
} 
 
/* A utility function to print solution */
void printSolution(int dist[][V]) 
{ 
    cout<<"The following matrix shows the shortest distances"
            " between every pair of vertices \n"; 
    for (int i = 0; i < V; i++) 
    { 
        for (int j = 0; j < V; j++) 
        { 
            if (dist[i][j] == INF) 
                cout<<"INF"<<"     "; 
            else
                cout<<dist[i][j]<<"     "; 
        } 
        cout<<endl; 
    } 
} 
 
// Driver code 
int main() 
{ 
    /* Let us create the following weighted graph 
            10 
    (0)------->(3) 
        |     /|\ 
    5 |     | 
        |     | 1 
    \|/     | 
    (1)------->(2) 
            3     */
    int graph[V][V] = { {0, 5, INF, 10}, 
                        {INF, 0, 3, INF}, 
                        {INF, INF, 0, 1}, 
                        {INF, INF, INF, 0} 
                    }; 
 
    // Print the solution 
    floydWarshall(graph); 
    return 0; 
} 
 
// This code is contributed by Mythri J L

C

// C Program for Floyd Warshall Algorithm
#include<stdio.h>
 
// Number of vertices in the graph
#define V 4
 
/* Define Infinite as a large enough 
  value. This value will be used
  for vertices not connected to each other */
#define INF 99999
 
// A function to print the solution matrix
void printSolution(int dist[][V]);
 
// Solves the all-pairs shortest path 
// problem using Floyd Warshall algorithm
void floydWarshall (int graph[][V])
{
    /* dist[][] will be the output matrix 
      that will finally have the shortest 
      distances between every pair of vertices */
    int dist[V][V], i, j, k;
 
    /* Initialize the solution matrix 
      same as input graph matrix. Or 
       we can say the initial values of 
       shortest distances are based
       on shortest paths considering no 
       intermediate vertex. */
    for (i = 0; i < V; i++)
        for (j = 0; j < V; j++)
            dist[i][j] = graph[i][j];
 
    /* Add all vertices one by one to 
      the set of intermediate vertices.
      ---> Before start of an iteration, we 
      have shortest distances between all
      pairs of vertices such that the shortest 
      distances consider only the
      vertices in set {0, 1, 2, .. k-1} as 
      intermediate vertices.
      ----> After the end of an iteration, 
      vertex no. k is added to the set of
      intermediate vertices and the set 
      becomes {0, 1, 2, .. k} */
    for (k = 0; k < V; k++)
    {
        // Pick all vertices as source one by one
        for (i = 0; i < V; i++)
        {
            // Pick all vertices as destination for the
            // above picked source
            for (j = 0; j < V; j++)
            {
                // If vertex k is on the shortest path from
                // i to j, then update the value of dist[i][j]
                if (dist[i][k] + dist[k][j] < dist[i][j])
                    dist[i][j] = dist[i][k] + dist[k][j];
            }
        }
    }
 
    // Print the shortest distance matrix
    printSolution(dist);
}
 
/* A utility function to print solution */
void printSolution(int dist[][V])
{
    printf ("The following matrix shows the shortest distances"
            " between every pair of vertices \n");
    for (int i = 0; i < V; i++)
    {
        for (int j = 0; j < V; j++)
        {
            if (dist[i][j] == INF)
                printf("%7s", "INF");
            else
                printf ("%7d", dist[i][j]);
        }
        printf("\n");
    }
}
 
// driver program to test above function
int main()
{
    /* Let us create the following weighted graph
            10
       (0)------->(3)
        |         /|\
      5 |          |
        |          | 1
       \|/         |
       (1)------->(2)
            3           */
    int graph[V][V] = { {0,   5,  INF, 10},
                        {INF, 0,   3, INF},
                        {INF, INF, 0,   1},
                        {INF, INF, INF, 0}
                      };
 
    // Print the solution
    floydWarshall(graph);
    return 0;
}

Java

// A Java program for Floyd Warshall All Pairs Shortest
// Path algorithm.
import java.util.*;
import java.lang.*;
import java.io.*;
 
 
class AllPairShortestPath
{
    final static int INF = 99999, V = 4;
 
    void floydWarshall(int graph[][])
    {
        int dist[][] = new int[V][V];
        int i, j, k;
 
        /* Initialize the solution matrix 
           same as input graph matrix.
           Or we can say the initial values 
           of shortest distances
           are based on shortest paths 
           considering no intermediate
           vertex. */
        for (i = 0; i < V; i++)
            for (j = 0; j < V; j++)
                dist[i][j] = graph[i][j];
 
        /* Add all vertices one by one 
           to the set of intermediate
           vertices.
          ---> Before start of an iteration, 
               we have shortest
               distances between all pairs 
               of vertices such that
               the shortest distances consider 
               only the vertices in
               set {0, 1, 2, .. k-1} as 
               intermediate vertices.
          ----> After the end of an iteration, 
                vertex no. k is added
                to the set of intermediate 
                vertices and the set
                becomes {0, 1, 2, .. k} */
        for (k = 0; k < V; k++)
        {
            // Pick all vertices as source one by one
            for (i = 0; i < V; i++)
            {
                // Pick all vertices as destination for the
                // above picked source
                for (j = 0; j < V; j++)
                {
                    // If vertex k is on the shortest path from
                    // i to j, then update the value of dist[i][j]
                    if (dist[i][k] + dist[k][j] < dist[i][j])
                        dist[i][j] = dist[i][k] + dist[k][j];
                }
            }
        }
 
        // Print the shortest distance matrix
        printSolution(dist);
    }
 
    void printSolution(int dist[][])
    {
        System.out.println("The following matrix shows the shortest "+
                         "distances between every pair of vertices");
        for (int i=0; i<V; ++i)
        {
            for (int j=0; j<V; ++j)
            {
                if (dist[i][j]==INF)
                    System.out.print("INF ");
                else
                    System.out.print(dist[i][j]+"   ");
            }
            System.out.println();
        }
    }
 
    // Driver program to test above function
    public static void main (String[] args)
    {
        /* Let us create the following weighted graph
           10
        (0)------->(3)
        |         /|\
        5 |          |
        |          | 1
        \|/         |
        (1)------->(2)
           3           */
        int graph[][] = { {0,   5,  INF, 10},
                          {INF, 0,   3, INF},
                          {INF, INF, 0,   1},
                          {INF, INF, INF, 0}
                        };
        AllPairShortestPath a = new AllPairShortestPath();
 
        // Print the solution
        a.floydWarshall(graph);
    }
}
 
// Contributed by Aakash Hasija

Python

# Python Program for Floyd Warshall Algorithm
 
# Number of vertices in the graph
V = 4
 
# Define infinity as the large 
# enough value. This value will be
# used for vertices not connected to each other
INF = 99999
 
# Solves all pair shortest path 
# via Floyd Warshall Algorithm
 
def floydWarshall(graph):
   
    """ dist[][] will be the output 
       matrix that will finally
        have the shortest distances 
        between every pair of vertices """
    """ initializing the solution matrix 
    same as input graph matrix
    OR we can say that the initial 
    values of shortest distances
    are based on shortest paths considering no 
    intermediate vertices """
 
    dist = list(map(lambda i: list(map(lambda j: j, i)), graph))
 
    """ Add all vertices one by one 
    to the set of intermediate
     vertices.
     ---> Before start of an iteration, 
     we have shortest distances
     between all pairs of vertices 
     such that the shortest
     distances consider only the 
     vertices in the set 
    {0, 1, 2, .. k-1} as intermediate vertices.
      ----> After the end of a 
      iteration, vertex no. k is
     added to the set of intermediate 
     vertices and the 
    set becomes {0, 1, 2, .. k}
    """
    for k in range(V):
 
        # pick all vertices as source one by one
        for i in range(V):
 
            # Pick all vertices as destination for the
            # above picked source
            for j in range(V):
 
                # If vertex k is on the shortest path from
                # i to j, then update the value of dist[i][j]
                dist[i][j] = min(dist[i][j],
                                 dist[i][k] + dist[k][j]
                                 )
    printSolution(dist)
 
 
# A utility function to print the solution
def printSolution(dist):
    print "Following matrix shows the shortest distances\
 between every pair of vertices"
    for i in range(V):
        for j in range(V):
            if(dist[i][j] == INF):
                print "%7s" % ("INF"),
            else:
                print "%7d\t" % (dist[i][j]),
            if j == V-1:
                print ""
 
 
# Driver program to test the above program
# Let us create the following weighted graph
"""
            10
       (0)------->(3)
        |         /|\
      5 |          |
        |          | 1
       \|/         |
       (1)------->(2)
            3           """
graph = [[0, 5, INF, 10],
         [INF, 0, 3, INF],
         [INF, INF, 0,   1],
         [INF, INF, INF, 0]
         ]
# Print the solution
floydWarshall(graph)
# This code is contributed by Mythri J L

C#

// A C# program for Floyd Warshall All 
// Pairs Shortest Path algorithm.
 
using System;
 
public class AllPairShortestPath
{
    readonly static int INF = 99999, V = 4;
 
    void floydWarshall(int[,] graph)
    {
        int[,] dist = new int[V, V];
        int i, j, k;
 
        // Initialize the solution matrix 
        // same as input graph matrix
        // Or we can say the initial 
        // values of shortest distances
        // are based on shortest paths 
        // considering no intermediate
        // vertex
        for (i = 0; i < V; i++) {
            for (j = 0; j < V; j++) {
                dist[i, j] = graph[i, j];
            }
        }
 
        /* Add all vertices one by one to 
        the set of intermediate vertices.
        ---> Before start of a iteration,
             we have shortest distances
             between all pairs of vertices
             such that the shortest distances
             consider only the vertices in
             set {0, 1, 2, .. k-1} as 
             intermediate vertices.
        ---> After the end of a iteration, 
             vertex no. k is added
             to the set of intermediate
             vertices and the set
             becomes {0, 1, 2, .. k} */
        for (k = 0; k < V; k++)
        {
            // Pick all vertices as source
            // one by one
            for (i = 0; i < V; i++)
            {
                // Pick all vertices as destination
                // for the above picked source
                for (j = 0; j < V; j++)
                {
                    // If vertex k is on the shortest
                    // path from i to j, then update
                    // the value of dist[i][j]
                    if (dist[i, k] + dist[k, j] < dist[i, j]) 
                    {
                        dist[i, j] = dist[i, k] + dist[k, j];
                    }
                }
            }
        }
 
        // Print the shortest distance matrix
        printSolution(dist);
    }
 
    void printSolution(int[,] dist)
    {
        Console.WriteLine("Following matrix shows the shortest "+
                        "distances between every pair of vertices");
        for (int i = 0; i < V; ++i)
        {
            for (int j = 0; j < V; ++j)
            {
                if (dist[i, j] == INF) {
                    Console.Write("INF ");
                } else {
                    Console.Write(dist[i, j] + " ");
                }
            }
             
            Console.WriteLine();
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        /* Let us create the following
           weighted graph
              10
        (0)------->(3)
        |         /|\
        5 |         |
        |         | 1
        \|/         |
        (1)------->(2)
             3             */
        int[,] graph = { {0, 5, INF, 10},
                        {INF, 0, 3, INF},
                        {INF, INF, 0, 1},
                        {INF, INF, INF, 0}
                        };
         
        AllPairShortestPath a = new AllPairShortestPath();
 
        // Print the solution
        a.floydWarshall(graph);
    }
}
 
// This article is contributed by 
// Abdul Mateen Mohammed

PHP

<?php
// PHP Program for Floyd Warshall Algorithm 
 
// Solves the all-pairs shortest path problem
// using Floyd Warshall algorithm 
function floydWarshall ($graph, $V, $INF) 
{ 
    /* dist[][] will be the output matrix 
    that will finally have the shortest 
    distances between every pair of vertices */
    $dist = array(array(0,0,0,0),
                  array(0,0,0,0),
                  array(0,0,0,0),
                  array(0,0,0,0));
 
    /* Initialize the solution matrix same 
    as input graph matrix. Or we can say the 
    initial values of shortest distances are 
    based on shortest paths considering no 
    intermediate vertex. */
    for ($i = 0; $i < $V; $i++) 
        for ($j = 0; $j < $V; $j++) 
            $dist[$i][$j] = $graph[$i][$j]; 
 
    /* Add all vertices one by one to the set 
    of intermediate vertices. 
    ---> Before start of an iteration, we have 
    shortest distances between all pairs of 
    vertices such that the shortest distances 
    consider only the vertices in set 
    {0, 1, 2, .. k-1} as intermediate vertices. 
    ----> After the end of an iteration, vertex 
    no. k is added to the set of intermediate
    vertices and the set becomes {0, 1, 2, .. k} */
    for ($k = 0; $k < $V; $k++) 
    { 
        // Pick all vertices as source one by one 
        for ($i = 0; $i < $V; $i++) 
        { 
            // Pick all vertices as destination 
            // for the above picked source 
            for ($j = 0; $j < $V; $j++) 
            { 
                // If vertex k is on the shortest path from 
                // i to j, then update the value of dist[i][j] 
                if ($dist[$i][$k] + $dist[$k][$j] < 
                                    $dist[$i][$j]) 
                    $dist[$i][$j] = $dist[$i][$k] +
                                    $dist[$k][$j]; 
            } 
        } 
    } 
 
    // Print the shortest distance matrix 
    printSolution($dist, $V, $INF); 
} 
 
/* A utility function to print solution */
function printSolution($dist, $V, $INF) 
{ 
    echo "The following matrix shows the " .
             "shortest distances between " . 
                "every pair of vertices \n"; 
    for ($i = 0; $i < $V; $i++) 
    { 
        for ($j = 0; $j < $V; $j++) 
        { 
            if ($dist[$i][$j] == $INF) 
                echo "INF " ; 
            else
                echo $dist[$i][$j], " ";
        } 
        echo "\n"; 
    } 
} 
 
// Driver Code
 
// Number of vertices in the graph 
$V = 4 ;
 
/* Define Infinite as a large enough 
value. This value will be used for
vertices not connected to each other */
$INF = 99999 ;
 
/* Let us create the following weighted graph 
        10 
(0)------->(3) 
    |     /|\ 
5 |     | 
    |     | 1 
\|/     | 
(1)------->(2) 
        3     */
$graph = array(array(0, 5, $INF, 10), 
               array($INF, 0, 3, $INF), 
               array($INF, $INF, 0, 1), 
               array($INF, $INF, $INF, 0)); 
 
// Print the solution 
floydWarshall($graph, $V, $INF); 
 
// This code is contributed by Ryuga
?>

Output:

Following matrix shows the shortest distances between every pair of vertices
      0      5      8      9
    INF      0      3      4
    INF    INF      0      1
    INF    INF    INF      0

Time Complexity: O(V^3)
The above program only prints the shortest distances. We can modify the solution to print the shortest paths also by storing the predecessor information in a separate 2D matrix.
Also, the value of INF can be taken as INT_MAX from limits.h to make sure that we handle maximum possible value. When we take INF as INT_MAX, we need to change the if condition in the above program to avoid arithmetic overflow.

#include 

#define INF INT_MAX
..........................
if ( dist[i][k] != INF && 
     dist[k][j] != INF && 
     dist[i][k] + dist[k][j] < dist[i][j]
    )
 dist[i][j] = dist[i][k] + dist[k][j];
...........................

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