<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]]:

        def shortest(source, destination, e):
            best_so_far = {source: 0}
            nearest = [(0, source)]
            best_path = dict()

            visited = set()

            while nearest:
                dist0, curr = heapq.heappop(nearest)
                if curr == destination:
                    break
                if curr in visited:
                    assert dist0 >= best_so_far[curr]
                    continue
                visited.add(curr)
                for (v, w) in e[curr]:
                    dist = dist0 + w
                    if v not in best_so_far or best_so_far[v] > dist:
                        best_so_far[v] = dist
                        heapq.heappush(nearest, (dist, v))
                        best_path[v] = curr
            
            return best_so_far[destination], best_path


        all_e = dict()
        e1 = [[] for _ in range(n)]
        e2 = [[] for _ in range(n)]

        M_2_000_000_000 = 2_000_000_000

        for (a, b, w) in edges:
            w1 = 1 if w < 0 else w
            e1[a].append([b, w1])
            e1[b].append([a, w1])

            # (original weight, index in e2[a], index in e2[b])
            all_e[(a, b)] = (w, len(e2[a]), len(e2[b]))

            w2 = M_2_000_000_000 if w < 0 else w
            e2[a].append([b, w2])
            e2[b].append([a, w2])
        

        
        z1, best_path = shortest(source, destination, e1)
        if target < z1:
            return []
        if target == z1:
            return [(a, b, abs(w)) for (a, b, w) in edges]

        # target > z1

        z2, _ = shortest(source, destination, e2)
        if target > z2:
            return []
        if target == z2:
            return [(a, b, M_2_000_000_000 if w < 0 else w) for (a, b, w) in edges]

        # z1 < target < z2
        # print(':::', z1, target, z2)
        
        # Find all flexible edges on the optimal path in graph z1.  This is the shortest 
        # we can get.  We start from graph z2 and shorten the fliexbile_edges_on_optimal_path
        # until we get the desired distance.

        v = destination
        flexible_edges_on_optimal_path = []
        bb = [v]
        while v != source:
            w = best_path[v]
            bb.append(w)
            if (v, w) in all_e and all_e[(v, w)][0] < 0:
                flexible_edges_on_optimal_path.append((v, w))
            elif (w, v) in all_e and all_e[(w, v)][0] < 0:
                flexible_edges_on_optimal_path.append((w, v))
            v = w

        # r[(a, b)] = length override for edge (a, b) if not M_2_000_000_000
        # We also modify e2 in place as we go.
        r = dict()

        too_long_by = z2 - target
        # print("...", z2, target, too_long_by, flexible_edges_on_optimal_path)
        for (v, w) in flexible_edges_on_optimal_path:
            # d1 = shortest(v, w, e2)[0]
            # d2 = max(1, d1 - too_long_by)
            # r[(v, w)] = d2
            (_, i, j) = all_e[(v, w)]
            e2[v][i][1] = 1
            e2[w][j][1] = 1
            r[(v, w)] = 1
            sh1 = shortest(source, destination, e2)[0]
            if sh1 <= target:
                e2[v][i][1] = 1 + (target - sh1)
                e2[w][j][1] = 1 + (target - sh1)
                r[(v, w)] = 1 + (target - sh1)
                break

        eee = [
            (a, b, M_2_000_000_000) if w < 0 and (a, b) not in r
            else
            (a, b, r[(a, b)]) if w < 0
            else
            (a, b, w)
            for (a, b, w) in edges
        ]

        return eee

        # eyyy = dict(
        #     (k, l)
        #     for (a, b, w) in eee
        #     for (k, l) in [((a, b), w), ((b, a), w)]
        # )

        # hhh = shortest(source, destination, e2)
        # # print('!!!', hhh[0], target)

        # if hhh[0] != target:
        #     v = destination
        #     bb = [v]

        #     while v != source:
        #         w = hhh[1][v]
        #         bb.append((w, eyyy[(v, w)]))
        #         v = w
        #     # print('???', bb)

        # return eee
        

21688