# Minimum Cost Graph

Given **N** nodes on a 2-D plane represented as **(x _{i}, y_{i})**. 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:**

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 **N ^{N}**.

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++ implentation 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 initilize 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` `unin(` `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)) { ` ` ` `unin(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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

1.41421

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

## Recommended Posts:

- Minimum Cost of Simple Path between two nodes in a Directed and Weighted Graph
- Minimum odd cost path in a matrix
- Minimum cost to connect all cities
- Minimum cost to merge numbers from 1 to N
- Minimum possible travel cost among N cities
- Find the minimum cost to cross the River
- Minimum cost to traverse from one index to another in the String
- Minimum spanning tree cost of given Graphs
- Minimum cost to reach the top of the floor by climbing stairs
- Find the minimum cost to reach destination using a train
- Choose atleast two elements from array such that their GCD is 1 and cost is minimum
- Minimum cost to reverse edges such that there is path between every pair of nodes
- Minimum Cost Path with Left, Right, Bottom and Up moves allowed
- Minimum cost to reach from the top-left to the bottom-right corner of a matrix
- Minimum cost to connect weighted nodes represented as array

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.