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Minimum Cost Graph

  • Difficulty Level : Medium
  • Last Updated : 25 Nov, 2021

Given N nodes on a 2-D plane represented as (xi, yi). The nodes are said to be connected if the manhattan distance between them is 1. You can connect two nodes that are not connected at the cost of euclidean distance between them. The task is to connect the graph such that every node has a path from any node with minimum cost.

Examples: 

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Input: N = 3, edges[][] = {{1, 1}, {1, 1}, {2, 2}, {3, 2}} 
Output: 1.41421 
Since (2, 2) and (2, 3) are already connected. 
So we try to connect either (1, 1) with (2, 2) 
or (1, 1) with (2, 3) but (1, 1) with (2, 2) 
yields the minimum cost.



Input: N = 3, edges[][] = {{1, 1}, {2, 2}, {3, 3}} 
Output: 2.82843

Approach: The brute force approach is to connect each node with every other node and similarly for the other N nodes but in the worst case the time complexity will be NN
The other way is to find the cost of every pair of vertices with the euclidean distance and those pairs which are connected will have the cost as 0
After knowing the cost of each pair we will apply the Kruskal Algorithm for the minimum spanning tree and it will yield the minimum cost for connecting the graph. Note that for Kruskal Algorithm, you have to have the knowledge of Disjoint Set Union (DSU).

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Max number of nodes given
const int N = 500 + 10;
 
// arr is the parent array
// sz is the size of the
// subtree in DSU
int arr[N], sz[N];
 
// Function to initialize the parent
// and size array for DSU
void initialize()
{
    for (int i = 1; i < N; ++i) {
        arr[i] = i;
        sz[i] = 1;
    }
}
 
// Function to return the
// parent of the node
int root(int i)
{
    while (arr[i] != i)
        i = arr[i];
    return i;
}
 
// Function to perform the
// merge operation
void Union(int a, int b)
{
    a = root(a);
    b = root(b);
 
    if (a != b) {
        if (sz[a] < sz[b])
            swap(a, b);
 
        sz[a] += sz[b];
        arr[b] = a;
    }
}
 
// Function to return the minimum cost required
double minCost(vector<pair<int, int> >& p)
{
 
    // Number of points
    int n = (int)p.size();
 
    // To store the cost of every possible pair
    // as { cost, {to, from}}.
    vector<pair<double, pair<int, int> > > cost;
 
    // Calculating the cost of every possible pair
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (i != j) {
 
                // Getting Manhattan distance
                int x = abs(p[i].first - p[j].first)
                        + abs(p[i].second - p[j].second);
 
                // If the distance is 1 the cost is 0
                // or already connected
                if (x == 1) {
                    cost.push_back({ 0, { i + 1, j + 1 } });
                    cost.push_back({ 0, { j + 1, i + 1 } });
                }
                else {
 
                    // Calculating the euclidean distance
                    int a = p[i].first - p[j].first;
                    int b = p[i].second - p[j].second;
                    a *= a;
                    b *= b;
                    double d = sqrt(a + b);
                    cost.push_back({ d, { i + 1, j + 1 } });
                    cost.push_back({ d, { j + 1, i + 1 } });
                }
            }
        }
    }
 
    // Krushkal Algorithm for Minimum
    // spanning tree
    sort(cost.begin(), cost.end());
 
    // To initialize the size and
    // parent array
    initialize();
 
    double ans = 0.00;
    for (auto i : cost) {
        double c = i.first;
        int a = i.second.first;
        int b = i.second.second;
 
        // If the parents are different
        if (root(a) != root(b)) {
            Union(a, b);
            ans += c;
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
 
    // Vector pairs of points
    vector<pair<int, int> > points = {
        { 1, 1 },
        { 2, 2 },
        { 2, 3 }
    };
 
    // Function calling and printing
    // the answer
    cout << minCost(points);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
     
// Max number of nodes given
static int N = 500 + 10;
 
static class pair
{
    double c;
    int first, second;
     
    pair(double c, int first, int second)
    {
        this.c = c;
        this.first = first;
        this.second = second;
    }
}
 
// arr is the parent array
// sz is the size of the
// subtree in DSU
static int[] arr = new int[N],
              sz = new int[N];
  
// Function to initialize the parent
// and size array for DSU
static void initialize()
{
    for(int i = 1; i < N; ++i)
    {
        arr[i] = i;
        sz[i] = 1;
    }
}
  
// Function to return the
// parent of the node
static int root(int i)
{
    while (arr[i] != i)
        i = arr[i];
         
    return i;
}
  
// Function to perform the
// merge operation
static void union(int a, int b)
{
    a = root(a);
    b = root(b);
  
    if (a != b)
    {
        if (sz[a] < sz[b])
        {
            int tmp = a;
            a = b;
            b = tmp;
        }
         
        sz[a] += sz[b];
        arr[b] = a;
    }
}
  
// Function to return the minimum
// cost required
static double minCost(int[][] p)
{
     
    // Number of points
    int n = (int)p.length;
  
    // To store the cost of every possible pair
    // as { cost, {to, from}}.
    ArrayList<pair> cost = new ArrayList<>();
  
    // Calculating the cost of every possible pair
    for(int i = 0; i < n; ++i)
    {
        for(int j = 0; j < n; ++j)
        {
            if (i != j)
            {
                 
                // Getting Manhattan distance
                int x = Math.abs(p[i][0] - p[j][0]) +
                        Math.abs(p[i][1] - p[j][1]);
  
                // If the distance is 1 the cost is 0
                // or already connected
                if (x == 1)
                {
                    cost.add(new pair( 0, i + 1,
                                          j + 1 ));
                    cost.add(new pair( 0, j + 1,
                                          i + 1 ));
                }
                else
                {
                     
                    // Calculating the euclidean
                    // distance
                    int a = p[i][0] - p[j][0];
                    int b = p[i][1] - p[j][1];
                    a *= a;
                    b *= b;
                     
                    double d = Math.sqrt(a + b);
                    cost.add(new pair(d, i + 1,
                                         j + 1 ));
                    cost.add(new pair(d, j + 1,
                                         i + 1));
                }
            }
        }
    }
  
    // Krushkal Algorithm for Minimum
    // spanning tree
    Collections.sort(cost, new Comparator<>()
    {
        public int compare(pair a, pair b)
        {
            if(a.c <= b.c)
                return -1;
            else
                return 1;
        }
    });
  
    // To initialize the size and
    // parent array
    initialize();
  
    double ans = 0.00;
    for(pair i : cost)
    {
        double c = i.c;
        int a = i.first;
        int b = i.second;
  
        // If the parents are different
        if (root(a) != root(b))
        {
            union(a, b);
            ans += c;
        }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Vector pairs of points
    int[][] points = { { 1, 1 },
                       { 2, 2 },
                       { 2, 3 }};
     
    // Function calling and printing
    // the answer
    System.out.format("%.5f", minCost(points));
}
}
 
// This code is contributed by offbeat
Output: 
1.41421

 

Time Complexity: O(N*N)
Auxiliary Space: O(N*N)




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