diff --git a/README.md b/README.md
index 79223cc..fe61d55 100644
--- a/README.md
+++ b/README.md
@@ -217,3 +217,4 @@ Fork this project and add your username + profile link to the [Census](https://g
  - [Deepanshu Dabas](https://github.com/deepanshu22)
  - [Soubhik Rakshit](https://github.com/soubh1k) 
  - [Jasen Wyatt](https://github.com/jasenwyatt)
+ - [Dev Goyal](https://github.com/devgoyal04)
diff --git a/algorithms/floyd_warshall.cpp b/algorithms/floyd_warshall.cpp
new file mode 100644
index 0000000..6d85357
--- /dev/null
+++ b/algorithms/floyd_warshall.cpp
@@ -0,0 +1,105 @@
+// 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;  
+}