Upside Down Dijkstra

Research note: I used the Codeforces statement and the official APAC problem analysis PDF. The PDF's key observation is that the bad heap order behaves like a naive DFS; the proof and explicit weight formula below are written independently.

Why DFS Appears

Suppose the algorithm first-pops vertex $u$ with heap key $D$. Since it was the maximum heap item, every old item in the heap has key at most $(D,u)$. Then the algorithm pushes $(D+w(u,v),v)$ for every neighbor $v$, and every edge weight is positive, so every new distance is strictly larger than $D$.

That means: if $u$ has any unvisited neighbor, the next first-visited vertex must be one of those neighbors. Some newly pushed visited vertices may be popped and skipped first, but they do not enter $S$.

If $u$ has no unvisited neighbor, then $u$ cannot produce a useful new heap item, so the process effectively backtracks to an older vertex that still has an unvisited neighbor. This is exactly the control flow of iterative DFS: keep a stack, move to an unvisited neighbor when possible, otherwise pop finished vertices. Dijkstra got flipped upside down and accidentally invented DFS. Happens to the best of us, apparently.

So existence is equivalent to this check:

1. $s_1$ must be $1$.
2. Start with stack $[1]$ and mark $1$ processed.
3. Before processing the next requested vertex $x$, pop stack tops whose neighbors are all already processed.
4. If the stack is empty, impossible.
5. Otherwise, $x$ must be adjacent to the current stack top. If not, impossible.
6. Mark $x$ processed and push it.

Efficient Validation

We need to know when a vertex has no unprocessed neighbors. Let

$rem[v] =$ number of neighbors of $v$ that are not processed yet.

Initially $rem[v]=deg(v)$. When a vertex $x$ becomes processed, every neighbor $y$ loses one unprocessed neighbor, so we do rem[y]--. Across the whole algorithm, each undirected edge causes two decrements total, so this part is linear.

For the adjacency test between the stack top and the next vertex, the solution stores all edge keys in a sorted array and uses binary search. That makes the implementation deterministic and still easily fast enough: $O((n+m) log m)$.

Constructing Weights

Let $pos[v]$ be the position of vertex $v$ in $S$. For every edge $(u,v)$, set

$w(u,v)=n-max(pos[u],pos[v])+1$.

Because positions are from $1$ to $n$, every weight is in $[1,n]$, safely inside the required $[1,10^9]$.

Now consider this edge from the endpoint that is processed earlier. If $pos[u]<pos[v]$, then

$w(u,v)=n-pos[v]+1$.

So among all currently unvisited neighbors of a processed vertex $u$, the neighbor with the smallest position in $S$ gets the largest weight. Since the next vertex of $S$ is globally the earliest unprocessed vertex, it is also the earliest unprocessed neighbor of the DFS stack top whenever the validation says we should move from that top.

Why This Construction Works

Induct on the number of vertices already first-popped.

Assume the algorithm has produced the correct prefix of $S$, and the DFS stack from validation is the active stack.

Let $x$ be the next vertex in $S$. After popping finished stack vertices, validation says $x$ is adjacent to the current stack top $p$.

First, entries created by vertices that are no longer on the stack cannot lead to unvisited vertices: those vertices were popped only after all their neighbors were processed. They may create skipped junk in the heap, but junk is junk.

Second, compare useful entries from different stack depths. Take an ancestor $a$ of $p$, and let $c$ be the child of $a$ on the stack path toward $p$. When $c$ was chosen after $a$, every still-unvisited neighbor $y$ of $a$ had $pos[y]>pos[c]$, so our formula gave

$w(a,c)>w(a,y)$.

Thus the key that moved from $a$ to $c$ was larger than any pending useful key from $a$. Distances only increase while going deeper down the stack, so useful entries from deeper stack vertices dominate useful entries from shallower ancestors. Therefore the deepest stack vertex with an unprocessed neighbor, namely $p$, controls the next first-visited vertex.

Finally, among unvisited neighbors of $p$, the desired $x$ has the smallest position in $S$. Our weight formula gives it the largest newly pushed key from $p$, so it is the next unvisited vertex popped. Ties between future neighbors cannot happen because their positions are distinct; ties involving already visited vertices only affect skipped heap pops.

So if the DFS validation passes, these weights force the broken Dijkstra to output exactly $S$. If validation fails, no weights can fix it, because the max-heap rule with positive weights already forces the DFS-stack behavior.

Complexity

Building adjacency and maintaining rem is $O(n+m)$. Sorting edge keys and binary searching adjacency checks gives $O((n+m) log m)$ time and $O(n+m)$ memory. With $n<=100000$ and $m<=200000$, that is completely fine.