<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 Graph {
    vector<vector<pair<int, int>>> adjList;
    int n;

public:
    Graph(int n) : n(n) {
        adjList.resize(n);
    }

    void formList(const vector<vector<int>>& edges) {
        for (const auto& edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int wt = edge[2];
            adjList[u].push_back({v, wt});
            adjList[v].push_back({u, wt});
        }
    }

    int dijkstra(const vector<vector<int>>& edges, int src, int dest) {
        formList(edges);
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;

        vector<long long> dist(n, 1e9);

        dist[src] = 0;
        pq.push({0, src});

        while (!pq.empty()) {
            auto [wt, node] = pq.top();  // Destructure properly here
            pq.pop();

            // No need to process if a shorter distance is already found
            if (wt > dist[node]) continue;

            for (const auto& [nbrNode, nbrWt] : adjList[node]) {
                if (dist[nbrNode] > wt + nbrWt) {
                    dist[nbrNode] = wt + nbrWt;
                    pq.push({dist[nbrNode], nbrNode});
                }
            }
        }

        return dist[dest];
    }
};
class Solution {
public:
    vector<vector<int>> helper(vector<vector<int>>& edges){
        for(auto&edge:edges){
            if(edge[2]==-1){
                edge[2] = 1e9;
            }
        }
        return edges;
    }
    static int getPriority(const vector<int>& edge, int src, int dest) {
        int u = edge[0], v = edge[1];

        // First priority: edges between src and dest
        if ((u == src && v == dest) || (u == dest && v == src)) {
            return 1;
        }

        // Second priority: edges involving src or dest but not both
        if (u == src || v == src || u == dest || v == dest) {
            return 2;
        }

        // Third priority: all other edges
        return 3;
    }
    vector<vector<int>> modifiedGraphEdges(int n, vector<vector<int>>& edges, int source, int destination, int target) {
        Graph g(n);
        sort(edges.begin(), edges.end(), [&](const vector<int>& a, const vector<int>& b) {
            int priorityA = getPriority(a, source, destination);
            int priorityB = getPriority(b, source, destination);
            
            if (priorityA != priorityB) {
                return priorityA < priorityB;
            }
            
            // If priorities are the same, sort based on weights (optional, but useful)
            return a[2] < b[2];  // You can also change the comparison here to fit your needs
        });

        vector<vector<int>>tempEdges = helper(edges);
        int currDist = g.dijkstra(tempEdges, source, destination);

        if(currDist < target)
            return {};
        if(currDist == target)
            return tempEdges;

        for(auto&edge:edges){
            if(edge[2] == 1e9){
                edge[2] = 1;
                int curr = g.dijkstra(edges, source, destination);
                if(curr == target)
                    return edges;

                if(curr < target){
                    edge[2] += (target - curr);
                    return edges;
                }
            }
        }
        return {};
    }
};

auto init = [](){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return '8';
}();
292302