C++ program to find out the maximum sum of a minimally connected graph


Suppose, we are given a minimally connected graph. That means removing any edge will make the graph disconnected. The graph has n vertices and the edges are given in an array 'edges'. There is also an array 'vertexValues' given to us that contain n integer values.

Now, we do the following −

  • We write a positive integer on each of the vertices and then try to calculate a score.

  • There is an edge connecting two vertices, we put the smaller value of the two vertices on the edges.

  • We calculate the score by adding all the edge values.

We have to find the maximum value that can be achieved by putting the values on the vertices. We have to print the maximum total value and the values to be written on the vertices.

So, if the input is like n = 6, edges = {{1, 2}, {2, 3}, {2, 4}, {4, 5}, {3, 6}}, vertexValues = {1, 2, 3, 4, 5, 6}, then the output will be 15, 3 1 2 4 5 6, because we can put the values on the vertices from 0 to n – 1 in the given way 3 1 2 4 5 6.

To solve this, we will follow these steps −

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])

Example

Let us see the following implementation to get better understanding −

#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;
}

Input

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

Output

15
3 1 2 4 5 6

Updated on: 25-Feb-2022

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