Last active
April 5, 2021 18:47
-
-
Save zSANSANz/071012052e531dcaaec9b6f74b87d1c4 to your computer and use it in GitHub Desktop.
Floyd Warshall Algorithm on C++
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
// 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 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); | |
} | |
/* A utility function to print solution */ | |
void printSolution(int dist[][V]) | |
{ | |
printf ("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; | |
} |
dude, are you trying to just show off someone that you know all of those algorithm by just Copy Pasting from other sources! XD
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Din't get the point of a direct copy-paste from GeeksForGeeks?
Check this out: https://www.geeksforgeeks.org/floyd-warshall-algorithm-dp-16/