Since interesting edges lie in the shortest path tree of 1 or n, some smarts like this can be used to optimize?

As expected, for each edge we will compute the minimum path from 1 to n and backwards, if this edge was reversed. Let's focus just on the path from 1 to n. First, we run dijkstra (or floyd warshall) to find the shortest path on the given graph, let this path be $$$(v_1, \dots , v_k)$$$, where $$$k \leq n$$$. We will then run 2 floyd warshall's: $$$D_{i,j}$$$ denotes the minimum path from $$$i$$$ to $$$j$$$ on the given graph, $$$D'_{i,j}$$$ denotes the minimum path from $$$i$$$ to $$$j$$$ on the given graph, but without any of the edges on the minimum path found by dijkstra.

For each edge $$$(u, v)$$$ with cost $$$c$$$, that is not on the minimum path: if we decide not to use it, then it is best to go on the minimum path we found (as it's just like removing the edge from the graph). If we decide to use it, then we can safely use the value $$$D_{1, v} + c + D_{u, n}$$$ (and compare it with the minimum path); even though $$$D_{1, v}$$$ might use the edge unflipped, if it did then the resulting value will be larger than the minimum path. Same with $$$D_{u, n}$$$.

For each edge $$$(v_i, v_{i+1})$$$ with cost $$$c$$$, on the minimum path: if we decide not to use it, then, if we view the path as one that begins at $$$v_1$$$, ends at $$$v_k$$$ and sometimes leaves the minimum path (then returns), then we can claim it leaves the minimum path exactly once (it can't leave less since the minimum path is not a path after the flip):

First, for all $$$x < y$$$, we won't leave the minimum path at $$$v_y$$$ and return at $$$v_x$$$; depending on whether $$$y \leq i, x \leq i < y, i < x$$$, we can get rid of one of these vertices (and this detour around the minimum path), and instead use a part of the minimum path (while the path still doesn't use the flipped edge). So now we know that all those jumps around the minimum path just cover contiguous, disjoint segments of edges in the minimum path. So any jump around a contiguous segment which doesn't contain the flipped edge can be replaced with that segment in the minimum path.

This yields the simple solution: for each $$$1 \leq l \leq i, i+1 \leq r \leq k$$$, we consider the path with cost $$$D_{1,v_l} + D'_{v_l, v_r} + D_{v_r, n}$$$. We do this in $$$\mathcal{O}(k^2)$$$ for each edge on the minimum path, so $$$\mathcal{O}(k^3)$$$.

And let's not forget, if we decide to use the flipped edge on the minimum path: well actually it will never be optimal: take any such path, and right before you use the flipped edge to go from $$$v_{i+1}$$$ to $$$v_i$$$, just instead use $$$D_{v,n}$$$, which is guaranteed to be shorter.

So it's a very light cubic complexity.

My submission (which is not meant to be readable, just for reassurance: https://oj.uz/submission/201505