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$ curl repovive.com/roadmaps/graph-theory/floyd-warshall-algorithm/implementation-shortest-routes-ii

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$ curl repovive.com/roadmaps/graph-theory/floyd-warshall-algorithm/implementation-shortest-routes-ii

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CSES 1672 Shortest Routes II - Implementation - Floyd-Warshall Algorithm | Graph Theory | Repovive

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Graph Theory37 sections · 1633 units81h total

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CSES 1672 Shortest Routes II - Implementation

(Using adjacency matrix)

Here is the solution:

INF = 10^18
dist = 2D array of size (n+1) x (n+1), all INF
for i from 1 to n:
    dist[i][i] = 0

for each edge (a, b, w):
    dist[a][b] = min(dist[a][b], w)
    dist[b][a] = min(dist[b][a], w)

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

for each query (a, b):
    if dist[a][b] >= INF:
        print -1
    else:
        print dist[a][b]

Use a large constant like $10^{18}$ for infinity to avoid overflow when adding.

This runs in $O(n^3)$ time and uses $O(n^2)$ space.