Dijkstra's Shortest-Path Algorithm

Question: doesn’t BFS already compute shortest paths in linear time?
Answer:yes,IF $l_e = 1$ for every edge e.
Question: why not just replace each edge e by directed path of $l_e$unit length edges.
Answer: blows up graph too much.
Solution: Dijkstra’s shortest path algorithm.
Dijkstra’s Algorithm
Initialize:
$\cdot$ X = [s] [vertices processed so far]
$\cdot$ A[s] = 0 [computed shortest path distances]
$\cdot$ B[s] = empty path[computed shortest paths]
B[s] only to help explanation.
Main Loop
$\cdot$ while $X \neq V$:
Main Loop cont’ d:
$\cdot$ among all edges $(v,w)\in E$
with $v \in X, w \notin X$，
pick the one that minimizes
$A[v] + l_{vw}$
[call it (v*, w*)]
$\cdot$ add w* to X
$\cdot$ A[$w^*$]= A[$v^*$] + $l_{v^*w^*}$
$\cdot$ B[$w^*$] = B[$v^*$]$u(v^*,w^*)$
Heap Operations
Recall: perform Insert,Extract-Min in O(log n) time.
(1) a perfectly balanced binary tree
(2)Heap property: at every node, key <= children’s keys
(3)extract-min by swapping up last leaf, bubbling down
(4) insert via bubbling up.
Also: will need ability to delete from middle of heap.(bubble up or bubble down)
Two Invariants
Invariant #1: elements in heap = vertices of V-X
Invariant #2: for $v \notin X$
Key[v] = smallest Dijstra greedy score of an edge(u,v) in E with u in X
if($+\infty$ if no such edges exist)
Point: by invariants,
Maintaining the Invariants
To maintain Invariant #2:
[that $\forall v \notin X$]: Key[v] = smallest Dijkstra greedy score of edge(u,v) with u in X.
When w extracted from heap(added to X)
$\cdot$for each edge(w,v) in E:
$\cdot \cdot$ if v in V-X (in heap)
$\cdot \cdot \cdot$ delete v from heap
$\cdot \cdot \cdot$ recompute key[v] = min{key[v],A[w] + $l_{wv}$}
$\cdot \cdot \cdot$ re-Insert v into heap.