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Data Structure

for each i from 1 to n:
    for each j from 1 to n:
        if i == j:
            dist[i][j] = 0
        else if there is an edge from i to j:
            dist[i][j] = weight of the edge from i to j
        else:
            dist[i][j] = infinity

for each k from 1 to n:
    for each i from 1 to n:
        for each j from 1 to n:
            dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])

The time complexity of the Floyd-Warshall algorithm is O(n^3), where n is the number of nodes in the graph. The space complexity is also O(n^3), since the algorithm requires a 2D matrix of size n x n to store the distances between all pairs of nodes.

      

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