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Maximize the sum of products of the degrees between any two vertices of the tree in C++
Given the task is to construct a tree with a given integer N such that, the sum of degree(x) * degree(y) for all ordered pairs (x,y) is maximum and x is not equal to y.
Input −N=5
Output −50
Explanation
1 \ 2 \ 3 \ 4 \ 5 Degree of 1st node = 1 Degree of 2nd node = 2 Degree of 3rd node = 2 Degree of 4th node = 2 Degree of 5th node = 1
Product of all degrees for all ordered pairs (x, y) −
1st node = 1*2 + 1*2 + 1*2 + 1*1 = 7 2nd node = 2*1 + 2*2 + 2*2 + 2*1 = 12 3rd node = 2*1 + 2*2 + 2*2 + 2*1 = 12 4th node = 2*1 + 2*2 + 2*2 + 2*1 = 12 5th node = 1*2 + 1*2 + 1*2 + 1*1 = 7 Total sum = 50
Input −N=7
Output −122
Approach used in the below program as follows
Sum of degrees of all nodes in a tree is − (2 * N) – 2. Here N=number of nodes in the tree. In order to maximize the sum, the number of leaf nodes have to be minimized.
Inside Max() function initialize int sum=0 and create nested loops initializing x=1 and y=1 having conditions x<N and y<N.
Inside the nested loops first check if(x==y), if so then add continue; statement
Else initialize int degree=2 and using an if statement check if(x==1 || x==n). If so then put degreeX=1. Then initialize int degree=2 and do the same for variable y
Finally, before closing the loops, update the sum variable by writing − sum = (degreeX + degreeY)
Example
#include <bits/stdc++.h>
using namespace std;
int Max(int N){
int sum = 0;
for (int x = 1; x <= N; x++){
for (int y = 1; y <= N; y++){
if (x == y)
continue;
//Initializing degree for node x to 2
int degreeX = 2;
//If x is the leaf node or the root node
if (x == 1 || x == N)
degreeX = 1;
//Initializing degree for node y to 2
int degreeY = 2;
//If y is the leaf node or the root node
if (y == 1 || y == N)
degreeY = 1;
//Updating sum
sum += (degreeX * degreeY);
}
}
return sum;
}
//Main function
int main(){
int N = 5;
cout << Max(N);
}Output
If we run the above code we will get the following output −
50
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