<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:
    import heapq
    from collections import defaultdict
    # if any fixed path is < target, return []
    # initialize -1 edges to 1, if none < target, return [] 
    # else (all fixed paths >= target), use dijkstra to choose path with smallest (number of -1s, distance) out of all that is < target, make it equal to target
    # set all other edges to max value.
    # worst case other path(s) have these exact 2 -1s, but since their paths longer it won't be min.
    # def modifiedGraphEdges(self, n: int, edges: List[List[int]], source: int, destination: int, target: int) -> List[List[int]]:
        # toChange = set([(min(u, v), max(u, v)) for u, v, w in edges if w == -1])
        # changes = defaultdict(list)
        # adj = defaultdict(set)
        # for u, v, w in edges:
        #     w = max(w, 1)
        #     adj[u].add((w, v))
        #     adj[v].add((w, u))
            
        # dist = [[math.inf, math.inf] for i in range(n)]
        # dist[source] = [0, 0]

        # # print(toChange)

        # pq = [((0, 0), source)]

        # candidates = []
        # bestChanges = dict()

        # while pq:
        #     # print(changes)
        #     (chs, d), curr = heapq.heappop(pq)
        #     if curr == destination and d <= target:
        #         if chs == 0:
        #             if d < target:
        #                 return []
        #             if d == target:
        #                 return [[u, v, ((2 * 10**9) if (min(u, v), max(u, v)) in toChange else w)] for u, v, w in edges]
        #             continue

        #         candidates.append((chs, d))
        #         bestChanges[(chs, d)] = changes[((chs, d), curr)]
        #         continue
            
        #     for w, v, in adj[curr]:
        #         new = (chs + (1 if (min(curr, v), max(curr, v)) in toChange else 0), d + w)
        #         if new[0] < dist[v][0] or new[1] < dist[v][1]:
        #             dist[v][0] = min(new[0], dist[v][0])
        #             dist[v][1] = min(new[1], dist[v][1])
        #             changes[(new, v)] = changes[((chs, d), curr)][:]
        #             if (min(curr, v), max(curr, v)) in toChange:
        #                 changes[(new, v)].append((min(curr, v), max(curr, v)))
        #                 print(changes[(new, v)])
        #             heapq.heappush(pq, (new, v))
        
        # ans = []

        # print(candidates)
        # print(bestChanges)

        # if not candidates:
        #     return []
                
        # minI = candidates.index(min(candidates))
        # bestD = min(candidates)[1]
        # bestC = set(bestChanges[(min(candidates))])


        # for u, v, w in edges:
        #     curr = (min(u,v), max(u,v))
            
        #     if curr in bestC and bestD != target:
        #         ans.append([u, v, target-bestD+1])
        #         bestD = target
        #         continue
            
        #     if curr in bestC:
        #         ans.append([u, v, 1])
        #         continue

        #     if curr in toChange:
        #         ans.append([u, v, 2*10**9])
        #         continue

        #     ans.append([u, v, w])
        
        # return ans


# if any fixed path is < target, return []
# initialize -1 edges to 1, if none < target, return []
# else (all fixed paths >= target), use dijkstra to choose path with smallest (number of -1s, distance) out of all that is < target, make it equal to target
# set all other edges to max value.
# worst case other path(s) have these exact 2 -1s, but since their paths longer it won't be min.
    def modifiedGraphEdges(self, n, edges, source, destination, target):
        toChange = set([(min(u, v), max(u, v)) for u, v, w in edges if w == -1])
        changes = defaultdict(list)
        adj = defaultdict(set)
        for u, v, w in edges:
            w = max(w, 1)
            adj[u].add((w, v))
            adj[v].add((w, u))

        dist = defaultdict(lambda: math.inf)
        dist[(source, 0)] = 0

        # print(toChange)

        pq = [((0, 0), source)]

        candidates = []
        bestChanges = dict()

        while pq:
            # print(len(pq))
            # print(changes)

            # print(changes)

            (chs, d), curr = heapq.heappop(pq)

            if chs > len(toChange):
                continue

            if curr == destination and d <= target:
                if chs == 0:
                    if d < target:
                        return []
                    if d == target:
                        return [[u, v, ((2 * 10 ** 9) if (min(u, v), max(u, v)) in toChange else w)] for u, v, w in
                                edges]
                    continue

                candidates.append((chs, d))
                bestChanges[(chs, d)] = changes[((chs, d), curr)]
                break

            for w, v, in adj[curr]:
                if (min(curr, v), max(curr, v)) in changes[(chs, d), curr]:
                    continue
                new = (chs + (1 if (min(curr, v), max(curr, v)) in toChange else 0), d + w)
                if new[1] < dist[(v, new[0])]:
                    dist[(v, new[0])] = new[1]
                    changes[(new, v)] = changes[((chs, d), curr)][:]
                    if (min(curr, v), max(curr, v)) in toChange:
                        changes[(new, v)].append((min(curr, v), max(curr, v)))
                        # print(changes[(new, v)])
                    heapq.heappush(pq, (new, v))

        ans = []

        # print(candidates)
        # print(bestChanges)

        if not candidates:
            return []

        minI = candidates.index(min(candidates))
        bestD = min(candidates)[1]
        bestC = set(bestChanges[(min(candidates))])

        for u, v, w in edges:
            curr = (min(u, v), max(u, v))

            if curr in bestC and bestD != target:
                ans.append([u, v, target - bestD + 1])
                bestD = target
                continue

            if curr in bestC:
                ans.append([u, v, 1])
                continue

            if curr in toChange:
                ans.append([u, v, 2 * 10 ** 9])
                continue

            ans.append([u, v, w])

        return ans


                
                
        
            








        
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