<p>You are given an <strong>undirected weighted</strong> <strong>connected</strong> graph containing <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>, and an integer array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> with weight <code>w<sub>i</sub></code>.</p>

<p>Some edges have a weight of <code>-1</code> (<code>w<sub>i</sub> = -1</code>), while others have a <strong>positive</strong> weight (<code>w<sub>i</sub> &gt; 0</code>).</p>

<p>Your task is to modify <strong>all edges</strong> with a weight of <code>-1</code> by assigning them <strong>positive integer values </strong>in the range <code>[1, 2 * 10<sup>9</sup>]</code> so that the <strong>shortest distance</strong> between the nodes <code>source</code> and <code>destination</code> becomes equal to an integer <code>target</code>. If there are <strong>multiple</strong> <strong>modifications</strong> that make the shortest distance between <code>source</code> and <code>destination</code> equal to <code>target</code>, any of them will be considered correct.</p>

<p>Return <em>an array containing all edges (even unmodified ones) in any order if it is possible to make the shortest distance from </em><code>source</code><em> to </em><code>destination</code><em> equal to </em><code>target</code><em>, or an <strong>empty array</strong> if it&#39;s impossible.</em></p>

<p><strong>Note:</strong> You are not allowed to modify the weights of edges with initial positive weights.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/18/graph.png" style="width: 300px; height: 300px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 5, edges = [[4,1,-1],[2,0,-1],[0,3,-1],[4,3,-1]], source = 0, destination = 1, target = 5
<strong>Output:</strong> [[4,1,1],[2,0,1],[0,3,3],[4,3,1]]
<strong>Explanation:</strong> The graph above shows a possible modification to the edges, making the distance from 0 to 1 equal to 5.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/18/graph-2.png" style="width: 300px; height: 300px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[0,1,-1],[0,2,5]], source = 0, destination = 2, target = 6
<strong>Output:</strong> []
<strong>Explanation:</strong> The graph above contains the initial edges. It is not possible to make the distance from 0 to 2 equal to 6 by modifying the edge with weight -1. So, an empty array is returned.
</pre>

<p><strong class="example">Example 3:</strong></p>

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/19/graph-3.png" style="width: 300px; height: 300px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[1,0,4],[1,2,3],[2,3,5],[0,3,-1]], source = 0, destination = 2, target = 6
<strong>Output:</strong> [[1,0,4],[1,2,3],[2,3,5],[0,3,1]]
<strong>Explanation:</strong> The graph above shows a modified graph having the shortest distance from 0 to 2 as 6.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code><font face="monospace">1 &lt;= edges.length &lt;= n * (n - 1) / 2</font></code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i&nbsp;</sub>&lt;&nbsp;n</code></li>
	<li><code><font face="monospace">w<sub>i</sub>&nbsp;= -1&nbsp;</font></code>or <code><font face="monospace">1 &lt;= w<sub>i&nbsp;</sub>&lt;= 10<sup><span style="font-size: 10.8333px;">7</span></sup></font></code></li>
	<li><code>a<sub>i&nbsp;</sub>!=&nbsp;b<sub>i</sub></code></li>
	<li><code>0 &lt;= source, destination &lt; n</code></li>
	<li><code>source != destination</code></li>
	<li><code><font face="monospace">1 &lt;= target &lt;= 10<sup>9</sup></font></code></li>
	<li>The graph is connected, and there are no self-loops or repeated edges</li>
</ul>

class Solution:
    def modifiedGraphEdges(self, n: int, edges: List[List[int]], source: int, destination: int, target: int) -> List[List[int]]:
        flexEdgeLookup = dict()
        g = [dict() for _ in range(n)]

        for i, (u, v, w) in enumerate(edges):
            if w == -1:
                w = 1
                if u < v:
                    flexEdgeLookup[u, v] = i
                else:
                    flexEdgeLookup[v, u] = i
            else:
                g[u][v] = w
                g[v][u] = w
        dists = dict()
        pq = []
        parent = dict()
        def runDjikstra():
            nonlocal g, dists, pq, parent
            dists.clear()
            dists[source] = 0
            pq.clear()
            pq.append((0, source))
            parent.clear()
            while pq:
                curDist, cur = heappop(pq)
                if dists[cur] == curDist:
                    if cur == destination:
                        break
                    else:
                        for nxt, w in g[cur].items():
                            w += curDist
                            if w <= target and (nxt not in dists or w < dists[nxt]):
                                parent[nxt] = cur
                                dists[nxt] = w
                                heappush(pq, (w, nxt))

        # Run without flexible edges
        runDjikstra()
        if (destination in dists) and dists[destination] < target:
            # shorter path already exists, no point adding more edges, no target path
            return []
        if (destination in dists) and dists[destination] == target:
            # Found the path exactly on target
            for i in flexEdgeLookup.values():
                # make all flexible edges useless, no need for them
                edges[i][2] = target + 1
            return edges

        # turn on all flex edges, each must have a least 1 cost
        for u, v in flexEdgeLookup:
            g[u][v] = 1
            g[v][u] = 1
        
        # Run Dijskstra with all those flex edges on
        runDjikstra()
        if (destination not in dists):
            # destination is still unreachable, no solution
            return []
        if dists[destination] == target:
            # destination is exactly reacheable, solution
            for i in flexEdgeLookup.values():
                edges[i][2] = 1
            return edges

        # turn off all edges
        for (u, v), i in flexEdgeLookup.items():
            g[u][v] = target + 1
            g[v][u] = target + 1
            edges[i][2] = target + 1
        

        while True:
            # for every Djikstra run, try to adjust the weight of one of the flex edges on the path
            cur = destination
            lastIndex = -1
            while cur != source:
                prev = parent[cur]
                if prev < cur:
                    u, v = prev, cur
                else:
                    u, v = cur, prev
                try:
                    lastIndex = flexEdgeLookup[u, v]
                    edges[lastIndex][2] = 1
                    g[v][u] = 1
                    g[u][v] = 1
                except KeyError:
                    pass
                cur = prev
            edges[lastIndex][2] += target - dists[destination]
            u, v = sorted(edges[lastIndex][:2])
            g[v][u] += target - dists[destination]
            g[u][v] += target - dists[destination]
            # remove this edge
            del flexEdgeLookup[u, v]
            # rerun Djiskstra to make sure no shorter path left behind
            runDjikstra()
            if dists[destination] == target:
                break

        return edges
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