<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) 
    {
        vector<vector<int>> edge(100,vector<int>(100,-2));
        for(int i=0;i<edges.size();++i)
            edge[edges[i][0]][edges[i][1]]=edge[edges[i][1]][edges[i][0]]=edges[i][2];
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq;
        pq.push({0,source});
        vector<int> dis(n,1e9+7);
        dis[source]=0;
        while(pq.size())
        {
            if(dis[pq.top().second]!=pq.top().first)
            {
                pq.pop();
                continue;
            }
            for(int i=0;i<n;++i)
            {
                if(edge[pq.top().second][i]<0)
                    continue;
                if(dis[i]>edge[pq.top().second][i]+dis[pq.top().second])
                {
                    dis[i]=edge[pq.top().second][i]+dis[pq.top().second];
                    pq.push({dis[i],i});
                }
            }
            pq.pop();
        }
        if(dis[destination]<target)
            return {};
        else if(dis[destination]==target)
        {
            for(int i=0;i<edges.size();++i)
                if(edges[i][2]==-1)
                    edges[i][2]=1e9+7;
            return edges;
        }
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>> pq2;
        for(int x=1;x<n;++x)
        {
            while(pq2.size())
                pq2.pop();
            vector<vector<int>> dp(x+1,vector<int>(n,1e9+7));
            vector<vector<pair<int,int>>> from(x+1,vector<pair<int,int>>(n,{-1,-1}));
            dp[0][source]=0;
            pq2.push({0,source,0});
            while(pq2.size())
            {
                if(dp[pq2.top()[2]][pq2.top()[1]]!=pq2.top()[0])
                {
                    pq2.pop();
                    continue;
                }
                for(int i=0;i<n;++i)
                {
                    if(edge[pq2.top()[1]][i]==-2)
                        continue;
                    else if(edge[pq2.top()[1]][i]==-1)
                    {
                        if(pq2.top()[2]<x && dp[pq2.top()[2]+1][i]>1+dp[pq2.top()[2]][pq2.top()[1]])
                        {
                            dp[pq2.top()[2]+1][i]=1+dp[pq2.top()[2]][pq2.top()[1]];
                            from[pq2.top()[2]+1][i]={pq2.top()[1],pq2.top()[2]};
                            pq2.push({dp[pq2.top()[2]+1][i],i,pq2.top()[2]+1});
                        }
                    }
                    else
                    {
                        if(dp[pq2.top()[2]][i]>edge[pq2.top()[1]][i]+dp[pq2.top()[2]][pq2.top()[1]])
                        {
                            dp[pq2.top()[2]][i]=edge[pq2.top()[1]][i]+dp[pq2.top()[2]][pq2.top()[1]];
                            from[pq2.top()[2]][i]={pq2.top()[1],pq2.top()[2]};
                            pq2.push({dp[pq2.top()[2]][i],i,pq2.top()[2]});
                        }
                    }
                }
                pq2.pop();
            }
            if(dp[x][destination]<=target)
            {
                int miss=target-dp[x][destination];
                for(int i=destination,lvl=x;i!=source;)
                {
                    if(lvl!=from[lvl][i].second)
                    {
                        int t=from[lvl][i].first;
                        for(int z=0;z<edges.size();++z)
                        {
                            if(edges[z][0]==i && edges[z][1]==t)
                            {
                                edges[z][2]=1+miss;
                                miss=0;
                                break;
                            }
                            if(edges[z][0]==t && edges[z][1]==i)
                            {
                                edges[z][2]=1+miss;
                                miss=0;
                                break;
                            }
                        }
                    }
                    int ni=from[lvl][i].first,nlvl=from[lvl][i].second;
                    i=ni,lvl=nlvl;
                }
                for(int z=0;z<edges.size();++z)
                    if(edges[z][2]==-1)
                        edges[z][2]=1e9+7;
                return edges;
            }
        }
        return {};
    }
};
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