Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
C++ program to construct graph with certain conditions
Suppose we have two numbers N and K. Consider there is an undirected graph with N elements. N vertices satisfies the following conditions −
The graph is simple and connected
Vertices are numbered from 1 to N
Let M be the number of edges in the graph. The edges are numbered from 1 to M. The length of the edge is 1. And Edge i connects vertex U[i] to vertex V[i].
There are exactly K pairs of vertices (i, j) where i < j, such that the shortest distance between them is 2.
If such graph exists, we have to construct that graph. Otherwise return -1.
So, if the input is like N = 5; K = 3, then the output will be
Steps
To solve this, we will follow these steps −
if k > (n - 1) * (n - 2) / 2, then: print -1 print ((n - 1) * (n - 2) / 2 - k + n - 1) for initialize i := 1, when i < n, update (increase i by 1), do: print pair (1, i + 1) count := (n - 1) * (n - 2) / 2 - k for initialize i := 2, when i <= n, update (increase i by 1), do: for initialize j := i + 1, when j <= n, update (increase j by 1), do: if count <= 0, then: return print pair (i, j) (decrease count by 1)
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
void solve(int n, int k){
if (k > (n - 1) * (n - 2) / 2){
cout << -1 << endl;
}
cout << (n - 1) * (n - 2) / 2 - k + n - 1 << '\n';
for (int i = 1; i < n; i++){
cout << 1 << ", " << i + 1 << '\n';
}
int count = (n - 1) * (n - 2) / 2 - k;
for (int i = 2; i <= n; i++){
for (int j = i + 1; j <= n; j++){
if (count <= 0){
return;
}
cout << i << ", " << j << '\n';
count--;
}
}
}
int main(){
int N = 5;
int K = 3;
solve(N, K);
}Input
5, 3
Output
7 1, 2 1, 3 1, 4 1, 5 2, 3 2, 4 2, 5
Updated on: 03-Mar-2022
428 Views
Advertisements