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Floyd Warshall Algorithm | DP-16
• Difficulty Level : Medium
• Last Updated : 28 Apr, 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 ``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][j] > (dist[i][k] + dist[k][j])``                    ``&& (dist[k][j] != INF``                        ``&& dist[i][k] != INF))``                    ``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` `// 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(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

 ` 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];
...........................```