c++ - Kruskal's Algorithm in O(n log n) -


i'm interested in learning how run kruskal's algorithm on o(n log n) in c++. have implemented algorithm solution runs on o(n^2), , code below.

i know should possible run kruskal's on o(n log n), i'm stymied how can done. appreciate , tips.

#include <vector> #include <algorithm> #include <set> using namespace std;  //sort weight bool sorting (vector<int> i, vector<int> j) {return i[2]<j[2];}  class submap { private:     set<int> finder; public:     submap(int x) {finder.insert(x);}     submap(set<int> x) {finder = x;}     set<int> pointreturn() {return finder;}     //function add new set current tree     void add(set<int> np) {         finder.insert(np.begin(), np.end());      }     void add(int n) {         finder.insert(n);     }     int size() {return int(finder.size());}      //find function returns true if value not found     bool find(int x) {         if (finder.find(x) == finder.end())             return true;         else             return false;     } };  class map { private:     vector<submap> submaps; public:     int find(int x) {         int finder = -1;         for(int = 0; < int(submaps.size()); i++) {             if(!submaps[i].find(x)) {                 finder = i;                 break;             }         }         return finder;     }     void newsubmap(int a, int b) {         set<int> nextset;         nextset.insert(a);         nextset.insert(b);         submaps.push_back(submap(nextset));     }     void addendum(int a, int index) {          submaps[index].add(a);     }      submap subfind(int i) {return submaps[i];}      void fuse(int index1, int index2) {         submaps[index1].add(submaps[index2].pointreturn());         vector<submap> nextmaps;         for(int = 0; < int(submaps.size()); i++) {             if (i != index2)                 nextmaps.push_back(submaps[i]);         }         submaps = nextmaps;     }     int size() {return submaps[0].size();} };  map kruskal (vector<vector<int>> &graph, int weight, int junct) {     //sort graph     sort(graph.begin(), graph.end(), sorting);     map currmap;      int usedweight = 0;     for(int i=0; i<graph.size(); i++) {         int = currmap.find(graph[i][0]);         int b = currmap.find(graph[i][1]);          //the boolean expression here false if both points in same submap         if(a != b || == -1) {              usedweight += graph[i][2];              //if neither point in map far             if(a == -1 && b == -1) {                 currmap.newsubmap(graph[i][0], graph[i][1]);             }              //if 1 point in map far             else if (a != -1 && b == -1) {                 currmap.addendum(graph[i][1], a);             }             else if (a == -1 && b != -1) {                 currmap.addendum(graph[i][0], b);             }              //if both points in map, different submaps             else {                 currmap.fuse(a, b);             }         }         //if first set in map spanning, algorithm done         if(currmap.size() == junct)             break;     }      return (currmap); }


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