Program C++ untuk mencari jumlah maksimum graf yang disambungkan secara minimum

N := 100
Define arrays seq and res of size N.
Define an array tp of size N.
ans := 0
Define a function dfs(), this will take p, q,
   res[p] := seq[c]
   if p is not equal to 0, then:
      ans := ans + seq[c]
   (decrease c by 1)
   for each value x in tp[p], do:
      if x is not equal to q, then:
         dfs(x, p)
for initialize i := 0, when i + 1 < n, update (increase i by 1), do:
   tmp := first value of edges[i]- 1
   temp := second value of edges[i] - 1
   insert temp at the end of tp[tmp]
   insert tmp at the end of tp[temp]
for initialize i := 0, when i < n, update (increase i by 1), do:
   seq[i] := vertexValues[i]
c := n - 1
sort the array seq
dfs(0, 0)
print(ans)
for initialize i := n - 1, when i >= 0, update (decrease i by 1), do:
   print(res[i])

#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
#define N 100
int seq[N], res[N];
vector<int> tp[N];
int ans = 0, c;

void dfs(int p, int q) {
   res[p] = seq[c];
   if(p != 0)
      ans += seq[c];
   c--;
   for(auto x : tp[p]) {
      if(x != q)
         dfs(x, p);
   }
}
void solve(int n, vector<pair<int,int>> edges, int vertexValues[]){
   for(int i = 0; i + 1 < n; i++) {
      int tmp = edges[i].first - 1;
      int temp = edges[i].second - 1;
      tp[tmp].push_back(temp);
      tp[temp].push_back(tmp);
   }
   for(int i = 0; i < n; i++)
      seq[i] = vertexValues[i];
   c = n - 1;
   sort(seq, seq + n);
   dfs(0, 0);
   cout << ans << endl;
   for(int i = n - 1; i >= 0; i--)
      cout << res[i] << " ";
   cout << endl;
}
int main() {
   int n = 6;
   vector<pair<int,int>> edges = {{1, 2}, {2, 3}, {2, 4}, {4, 5},{3, 6}};
   int vertexValues[] = {1, 2, 3, 4, 5, 6};
   solve(n, edges, vertexValues);
   return 0;
}

6, {{1, 2}, {2, 3}, {2, 4}, {4, 5}, {3, 6}}, {1, 2, 3, 4, 5, 6}

15
3 1 2 4 5 6