ronith
9/6/2018 - 10:00 AM

Floyd Warshall Algorithm

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].
// https://www.geeksforgeeks.org/?p=19772
#include<iostream>
using namespace std;

#define inf INT_MAX

void floydWarshall ( int v, int graph[][4]) {
    int dist[v][v];
    for (int i=0;i<v;i++)
        for(int j=0;j<v;j++)
            dist[i][j] = graph[i][j];

    for (int i=0;i<v;i++) {
        for(int j=0;j<v;j++) {
            for (int k=0; k<v;k++) {
                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];
            }
        }
    }
    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]);
        }
        cout<< "\n";
    }
}

 int main() {
     int v=4;
    int graph[v][4] = { {0,   5,  inf, 10},
                        {inf, 0,   3, inf},
                        {inf, inf, 0,   1},
                        {inf, inf, inf, 0}};
    floydWarshall(v,graph);
}