<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 {
public:
    vector<vector<int>> modifiedGraphEdges(int n, vector<vector<int>>& edges, int source, int destination, int target) {
        #pragma GCC optimize ("Ofast")
        ios_base::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        typedef pair<int, int> PII;
        vector<vector<PII>> adjLists(n);
        deque<vector<int>> newEdges;
        int len = edges.size();
        for (int i = 0; i < len; i++) {
            int v1 = edges[i][0], v2 = edges[i][1], w = edges[i][2];
            if (w == -1) {
                newEdges.push_back(edges[i]);
                newEdges.back()[2] = i;           // change -1 into edge index
            }
            else {
                adjLists[v1].push_back({v2, w});
                adjLists[v2].push_back({v1, w});
            }
        }
        auto dijkstra = [&] (int source) {
            // Caution: minDistance should be reinitilaized every time!!
            vector<int> minDistance(n, INT_MAX);
            minDistance[source] = 0;
            priority_queue<PII, vector<PII>, greater<PII>> PQ;
            PQ.push({0, source});
            while (!PQ.empty()) {
                auto [cost, curr] = PQ.top();
                PQ.pop();
                for (auto edge: adjLists[curr]) {
                    auto [nei, weight] = edge;
                    if (minDistance[nei] > weight + cost) {
                        minDistance[nei] = weight + cost;
                        PQ.push({weight+cost, nei});
                    }
                }
            }
            return minDistance[destination];
        };
        int currDist = dijkstra(source);
        // if current distance is less than target distance, 
        // then it's impossible to modify graph to satisfy target distance.
        if (currDist < target)
            return {};
        bool matchTarget = (currDist == target);
        while (!newEdges.empty()) {
            auto newedge = newEdges.front();
            newEdges.pop_front();
            int v1 = newedge[0], v2 = newedge[1], idx = newedge[2];
            // caution: do not set to INT_MAX to avoid integer overflow
            int newWeight = (matchTarget) ? 1e9 : 1;
            edges[idx][2] = newWeight;
            adjLists[v1].push_back({v2, newWeight});
            adjLists[v2].push_back({v1, newWeight});
            if (!matchTarget) {
                int currDist = dijkstra(source);
                if (currDist <= target) {
                    matchTarget = true;
                    edges[idx][2] += target - currDist;
                }
            }
        }
        if (!matchTarget)
            return {};
        return edges;
    }
};
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