After each iteration of the while loop, the set \(S\) will contain vertices \(v\) whose \(d[v]\) and \(p[v]\) values are known to be correct.

for all v ∈ V do { d[v] ← ∞;  p[v] ← v }
d[s] ← 0;  p[s] ← s
S ← ∅
while there is a vertex v ∈ V\S with d[v] < ∞ do {
    v ← vertex in V\S with minimum d value
    S ← S ∪ {v}
    for each edge (v,w) do {
        if w is not in S then {
            if d[v] + l[v,w] < d[w] then {
                d[w] ← d[v] + l[v,w]
                p[w] ← v
            }
        }
    }
}

1. \(d[v]\) equals the length of a shortest path from \(s\) to \(v\) if \(v\) is reachable from \(s\); it equals \(\infty\) otherwise.
2. the tree rooted at \(s\) as defined by the parent pointers \(p[v]\) is a shortest-path tree. (The parent of \(s\) is \(s\) itself.)

• finds \(v\) by deleting the minimum key in \(Q\)
• scans the adjacency list of \(v\) to update keys; if \(d[w]\) decreases, deletes \(w\) from \(Q\) and then reinserts it in \(Q\) with the new key

• implementing \(Q\) as a binary heap or red-black tree gives time \(O(m\log n)\)
• implementing \(Q\) as an array gives tive \(O(n^2)\)
• implementing \(Q\) as a Fibonacci heap gives time \(O(m+n\log n)\), the best bound for implementing Dijkstra's algorithm

• for all \(v\in S\), d[v] is the correct distance of of \(v\) from \(s\)
• \(p[v]\) for all \(v\in S\) comprises the edges of a shortest-path tree spanning \(S\)