<p>Given a weighted undirected connected graph with <code>n</code>&nbsp;vertices numbered from <code>0</code> to <code>n - 1</code>,&nbsp;and an array <code>edges</code>&nbsp;where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>, weight<sub>i</sub>]</code> represents a bidirectional and weighted edge between nodes&nbsp;<code>a<sub>i</sub></code>&nbsp;and <code>b<sub>i</sub></code>. A minimum spanning tree (MST) is a subset of the graph&#39;s edges that connects all vertices without cycles&nbsp;and with the minimum possible total edge weight.</p>

<p>Find <em>all the critical and pseudo-critical edges in the given graph&#39;s minimum spanning tree (MST)</em>. An MST edge whose deletion from the graph would cause the MST weight to increase is called a&nbsp;<em>critical edge</em>. On&nbsp;the other hand, a pseudo-critical edge is that which can appear in some MSTs but not all.</p>

<p>Note that you can return the indices of the edges in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/06/04/ex1.png" style="width: 259px; height: 262px;" /></p>

<pre>
<strong>Input:</strong> n = 5, edges = [[0,1,1],[1,2,1],[2,3,2],[0,3,2],[0,4,3],[3,4,3],[1,4,6]]
<strong>Output:</strong> [[0,1],[2,3,4,5]]
<strong>Explanation:</strong> The figure above describes the graph.
The following figure shows all the possible MSTs:
<img alt="" src="https://assets.leetcode.com/uploads/2020/06/04/msts.png" style="width: 540px; height: 553px;" />
Notice that the two edges 0 and 1 appear in all MSTs, therefore they are critical edges, so we return them in the first list of the output.
The edges 2, 3, 4, and 5 are only part of some MSTs, therefore they are considered pseudo-critical edges. We add them to the second list of the output.
</pre>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/06/04/ex2.png" style="width: 247px; height: 253px;" /></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[0,1,1],[1,2,1],[2,3,1],[0,3,1]]
<strong>Output:</strong> [[],[0,1,2,3]]
<strong>Explanation:</strong> We can observe that since all 4 edges have equal weight, choosing any 3 edges from the given 4 will yield an MST. Therefore all 4 edges are pseudo-critical.
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= edges.length &lt;= min(200, n * (n - 1) / 2)</code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt; n</code></li>
	<li><code>1 &lt;= weight<sub>i</sub>&nbsp;&lt;= 1000</code></li>
	<li>All pairs <code>(a<sub>i</sub>, b<sub>i</sub>)</code> are <strong>distinct</strong>.</li>
</ul>

class DisjointSet {
    vector<int> rank, parent; 
public: 
    DisjointSet(int n) {
        rank.resize(n+1, 0); 
        parent.resize(n+1);
        for(int i = 0;i<=n;i++) {
            parent[i] = i; 
        }
    }

    int findUPar(int node) {
        if(node == parent[node])
            return node; 
        return parent[node] = findUPar(parent[node]); 
    }

    void unionByRank(int u, int v) {
        int ulp_u = findUPar(u); 
        int ulp_v = findUPar(v); 
        if(ulp_u == ulp_v) return; 
        if(rank[ulp_u] < rank[ulp_v]) {
            parent[ulp_u] = ulp_v; 
        }
        else if(rank[ulp_v] < rank[ulp_u]) {
            parent[ulp_v] = ulp_u; 
        }
        else {
            parent[ulp_v] = ulp_u; 
            rank[ulp_u]++; 
        }
    }
    bool isConnected(int u,int v){
        int u_par=findUPar(u);
        int v_par=findUPar(v);
        return (u_par==v_par);
    }
}; 
class Solution {
private:
    int calMST(vector<vector<int>>& v,int n,int leave){
        DisjointSet ds(n);
        int sum=0;
        for(auto it:v){
            int u=it[1];
            int v=it[2];
            int w=it[0];
            int ind=it[3];
            if(ind==leave){
                continue;
            }
            if(!ds.isConnected(u,v)){
                ds.unionByRank(u,v);
                sum+=w;
            }
        }
        for(int i=0;i<n;i++){
            if(!ds.isConnected(0,i)){
                return 1e9;
            }
        }
        return sum;
    }
    int isPossibletoTake(vector<vector<int>>& v,int n,vector<int>& take){
        DisjointSet ds(n);
        int sum=take[0];
        ds.unionByRank(take[1],take[2]);
        int leave=take[3];
        for(auto it:v){
            int u=it[1];
            int v=it[2];
            int w=it[0];
            int ind=it[3];
            if(ind==leave){
                continue;
            }
            if(!ds.isConnected(u,v)){
                ds.unionByRank(u,v);
                sum+=w;
            }
        }
        for(int i=0;i<n;i++){
            if(!ds.isConnected(0,i)){
                return 1e9;
            }
        }
        return sum;
    }
public:
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        vector<vector<int>> v;
        int m=edges.size();
        for(int i=0;i<m;i++){
            v.push_back({edges[i][2],edges[i][0],edges[i][1],i});
        }
        sort(v.begin(),v.end());
        int MSTw=calMST(v,n,-1);
        cout<<MSTw<<endl;
        vector<int> critical,pseudo;
        for(int i=0;i<m;i++){
            int w=calMST(v,n,v[i][3]);
            cout<<"FIRST "<<v[i][3]<<" "<<w<<endl;
            if(w>MSTw){
                critical.push_back(v[i][3]);
            }
            else if(isPossibletoTake(v,n,v[i])==MSTw && w==MSTw){
                pseudo.push_back(v[i][3]);
            }
        }
        return {critical,pseudo};
    }
};
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