<p>You are given an undirected graph (the <strong>&quot;original graph&quot;</strong>) with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. You decide to <strong>subdivide</strong> each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge.</p>

<p>The graph is given as a 2D array of <code>edges</code> where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, cnt<sub>i</sub>]</code> indicates that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> in the original graph, and <code>cnt<sub>i</sub></code> is the total number of new nodes that you will <strong>subdivide</strong> the edge into. Note that <code>cnt<sub>i</sub> == 0</code> means you will not subdivide the edge.</p>

<p>To <strong>subdivide</strong> the edge <code>[u<sub>i</sub>, v<sub>i</sub>]</code>, replace it with <code>(cnt<sub>i</sub> + 1)</code> new edges and <code>cnt<sub>i</sub></code> new nodes. The new nodes are <code>x<sub>1</sub></code>, <code>x<sub>2</sub></code>, ..., <code>x<sub>cnt<sub>i</sub></sub></code>, and the new edges are <code>[u<sub>i</sub>, x<sub>1</sub>]</code>, <code>[x<sub>1</sub>, x<sub>2</sub>]</code>, <code>[x<sub>2</sub>, x<sub>3</sub>]</code>, ..., <code>[x<sub>cnt<sub>i</sub>-1</sub>, x<sub>cnt<sub>i</sub></sub>]</code>, <code>[x<sub>cnt<sub>i</sub></sub>, v<sub>i</sub>]</code>.</p>

<p>In this <strong>new graph</strong>, you want to know how many nodes are <strong>reachable</strong> from the node <code>0</code>, where a node is <strong>reachable</strong> if the distance is <code>maxMoves</code> or less.</p>

<p>Given the original graph and <code>maxMoves</code>, return <em>the number of nodes that are <strong>reachable</strong> from node </em><code>0</code><em> in the new graph</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/08/01/origfinal.png" style="width: 600px; height: 247px;" />
<pre>
<strong>Input:</strong> edges = [[0,1,10],[0,2,1],[1,2,2]], maxMoves = 6, n = 3
<strong>Output:</strong> 13
<strong>Explanation:</strong> The edge subdivisions are shown in the image above.
The nodes that are reachable are highlighted in yellow.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> edges = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], maxMoves = 10, n = 4
<strong>Output:</strong> 23
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> edges = [[1,2,4],[1,4,5],[1,3,1],[2,3,4],[3,4,5]], maxMoves = 17, n = 5
<strong>Output:</strong> 1
<strong>Explanation:</strong> Node 0 is disconnected from the rest of the graph, so only node 0 is reachable.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= edges.length &lt;= min(n * (n - 1) / 2, 10<sup>4</sup>)</code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= u<sub>i</sub> &lt; v<sub>i</sub> &lt; n</code></li>
	<li>There are <strong>no multiple edges</strong> in the graph.</li>
	<li><code>0 &lt;= cnt<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= maxMoves &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= n &lt;= 3000</code></li>
</ul>

class Solution {
public:
    int inf=1e7;
    #define ff first
    #define ss second
    int reachableNodes(vector<vector<int>>& a, int k, int n) {
        vector<vector<pair<int,int>>>adj(n);
        for(int i=0;i<a.size();i++){
            adj[a[i][0]].push_back({a[i][1],a[i][2]+1});
            adj[a[i][1]].push_back({a[i][0],a[i][2]+1});
        }
        
        using t = pair<int,int> ;
        vector<int>dp(n,inf);
        dp[0]=0;
        priority_queue<t,vector<t>,greater<t>>pq;
        pq.push({0,0});
        while(!pq.empty()){
            t p=pq.top();
            pq.pop();
            int v=p.ss,d=p.ff;
            if(d!=dp[v]){
                continue;
            }
            for(auto e:adj[v]){
                int u=e.ff,w=e.ss;
                if(dp[v]!=inf and d+w<=k and dp[u]>dp[v]+w){
                    dp[u] = dp[v]+w;
                    pq.push({dp[u],u});
                }
            }
        }
        set<int> s;
        int ans=0;
        s.insert(0);
        for(int i=0;i<a.size();i++){
            int v=a[i][0],u=a[i][1],nodes=a[i][2];
            int left = 0;
            if(dp[v]==inf and dp[u]==inf){
                continue;
            }
            if(dp[v]!=inf){
                left+=k-dp[v];
                s.insert(v);
            }
            if(dp[u]!=inf){
                left+=k-dp[u];
                s.insert(u);
            }
            ans+=min(left,nodes);

        }
        ans+=s.size();
        return ans;
    }
};
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