<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>

import heapq

class Solution:
    def modifiedGraphEdges(
        self, n: int, edges: List[List[int]], source: int, destination: int, target: int
    ) -> List[List[int]]:

        adjacency_matrix = defaultdict(set)
        for i, [a, b, w] in enumerate(edges):
            adjacency_matrix[a].add((b, w, i))
            adjacency_matrix[b].add((a, w, i))

        (dist, swaps) = traverse3(adjacency_matrix, source, destination, target)

        if len(swaps) == 0:
            if dist < target:
                return []
            if dist == target:
                return [[a, b, (w > 0) * w + (w < 0) * target] for [a, b, w] in edges]

        last_i = swaps.pop()

        def reweight(i, w):
            if i == last_i:
                return target - dist + 1
            if i in swaps:
                return 1
            if w < 0:
                return target
            return w

        # print(swapdict)
        return [[a, b, reweight(i, w)] for i, [a, b, w] in enumerate(edges)]


def dijkstra(adjacency_matrix, source):
    # do dijkstra and output full distance matrix
    to_visit = [(source,0)]
    distance = dict()
    while to_visit:
        to_visit.sort(key=lambda t:-t[1])
        curr, dist = to_visit.pop()
        for adj, adj_dist, ind in adjacency_matrix[curr]:
            if adj not in visited:
                new_dist = dist + abs(adj_dist)
                distance[adj] = new_dist
                to_visit.append((adj, new_dist))
    return distance
            


def traverse3(adjacency_matrix, source, destination, target):
    # do a dijkstra type algorithm
    #   we want to find the path to the dest. with the fewest # of unknown edges
    #   while not going over the target distance.

    # traverse2 but now with heaps
    to_visit = [(0,0, source, set(), set())]
    heapq.heapify(to_visit)
    while len(to_visit)>0:
        (mods_no, dist, curr, hist, mods) = heapq.heappop(to_visit)
        if curr == destination:
            return (dist, mods)
        hist.add(curr)
        for adj, adj_dist, ind in adjacency_matrix[curr]:
            new_dist = dist + abs(adj_dist)
            if new_dist <= target:
                new_mods = mods.copy()
                new_hist = hist.copy()
                if adj_dist == -1:
                    new_mods.add(ind)
                if adj not in hist:
                    heapq.heappush(to_visit, (len(new_mods), new_dist, adj, new_hist, new_mods))
    return (0, set())


def traverse2(adjacency_matrix, source, destination, target):
    # do a dijkstra type algorithm
    #   we want to find the path to the dest. with the fewest # of unknown edges
    #   while not going over the target distance.

    to_visit = [(source, 0, set(), set())]
    while to_visit:
        to_visit.sort(key=lambda t: (-len(t[3]), -t[1]))
        [curr, dist, hist, mods] = to_visit.pop()
        if curr == destination:
            return (dist, mods)
        hist.add(curr)
        for adj, adj_dist, ind in adjacency_matrix[curr]:
            new_dist = dist + abs(adj_dist)
            if new_dist <= target:
                new_mods = mods.copy()
                new_hist = hist.copy()
                if adj_dist == -1:
                    new_mods.add(ind)
                if adj not in hist:
                    to_visit.append([adj, new_dist, new_hist, new_mods])
    return (0, set())

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