A board of length m and width n is given, we need to break this board into m*n squares such that cost of breaking is minimum. cutting cost for each edge will be given for the board. In short we need to choose such a sequence of cutting such that cost is minimized.
Examples:
For above board optimal way to cut into square is: Total minimum cost in above case is 42. It is evaluated using following steps. Initial Value : Total_cost = 0 Total_cost = Total_cost + edge_cost * total_pieces Cost 4 Horizontal cut Cost = 0 + 4*1 = 4 Cost 4 Vertical cut Cost = 4 + 4*2 = 12 Cost 3 Vertical cut Cost = 12 + 3*2 = 18 Cost 2 Horizontal cut Cost = 18 + 2*3 = 24 Cost 2 Vertical cut Cost = 24 + 2*3 = 30 Cost 1 Horizontal cut Cost = 30 + 1*4 = 34 Cost 1 Vertical cut Cost = 34 + 1*4 = 38 Cost 1 Vertical cut Cost = 38 + 1*4 = 42
This problem can be solved using greedy approach, If total cost is denoted by S, then S = a1w1 + a2w2 … + akwk, where wi is a cost of certain edge cutting and ai is corresponding coefficient, The coefficient ai is determined by the total number of cuts we have competed using edge wi at the end of the cutting process. Notice that sum of the coefficients are always constant, hence we want to find a distribution of ai obtainable such that S is minimum. To do so we perform cuts on highest cost edge as early as possible, which will reach to optimal S. If we encounter several edges having the same cost, we can cut any one of them first.
Below is the solution using above approach, first we sorted the edge cutting costs in reverse order, then we loop in them from higher cost to lower cost building our solution. Each time we choose an edge, counter part count is incremented by 1, which is to be multiplied each time with corresponding edge cutting cost.
Notice below used sort method, sending greater() as 3rd argument to sort method sorts number in non-increasing order, it is predefined function of the library.
C++
// C++ program to divide a board into m*n squares #include <bits/stdc++.h> using namespace std; // method returns minimum cost to break board into // m*n squares int minimumCostOfBreaking(int X[], int Y[], int m, int n) { int res = 0; // sort the horizontal cost in reverse order sort(X, X + m, greater<int>()); // sort the vertical cost in reverse order sort(Y, Y + n, greater<int>()); // initialize current width as 1 int hzntl = 1, vert = 1; // loop untill one or both cost array are processed int i = 0, j = 0; while (i < m && j < n) { if (X[i] > Y[j]) { res += X[i] * vert; // increase current horizontal part count by 1 hzntl++; i++; } else { res += Y[j] * hzntl; // increase current vertical part count by 1 vert++; j++; } } // loop for horizontal array, if remains int total = 0; while (i < m) total += X[i++]; res += total * vert; // loop for vertical array, if remains total = 0; while (j < n) total += Y[j++]; res += total * hzntl; return res; } // Driver code to test above methods int main() { int m = 6, n = 4; int X[m-1] = {2, 1, 3, 1, 4}; int Y[n-1] = {4, 1, 2}; cout << minimumCostOfBreaking(X, Y, m-1, n-1); return 0; } |
Java
// Java program to divide a // board into m*n squares import java.util.Arrays; import java.util.Collections; class GFG { // method returns minimum cost to break board into // m*n squares static int minimumCostOfBreaking(Integer X[], Integer Y[], int m, int n) { int res = 0; // sort the horizontal cost in reverse order Arrays.sort(X, Collections.reverseOrder()); // sort the vertical cost in reverse order Arrays.sort(Y, Collections.reverseOrder()); // initialize current width as 1 int hzntl = 1, vert = 1; // loop untill one or both // cost array are processed int i = 0, j = 0; while (i < m && j < n) { if (X[i] > Y[j]) { res += X[i] * vert; // increase current horizontal // part count by 1 hzntl++; i++; } else { res += Y[j] * hzntl; // increase current vertical // part count by 1 vert++; j++; } } // loop for horizontal array, // if remains int total = 0; while (i < m) total += X[i++]; res += total * vert; // loop for vertical array, // if remains total = 0; while (j < n) total += Y[j++]; res += total * hzntl; return res; } // Driver program public static void main(String arg[]) { int m = 6, n = 4; Integer X[] = {2, 1, 3, 1, 4}; Integer Y[] = {4, 1, 2}; System.out.print(minimumCostOfBreaking(X, Y, m-1, n-1)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python program to divide a board into m*n squares # Method returns minimum cost to # break board into m*n squares def minimumCostOfBreaking(X, Y, m, n): res = 0 # sort the horizontal cost in reverse order X.sort(reverse = True) # sort the vertical cost in reverse order Y.sort(reverse = True) # initialize current width as 1 hzntl = 1; vert = 1 # loop untill one or both # cost array are processed i = 0; j = 0 while (i < m and j < n): if (X[i] > Y[j]): res += X[i] * vert # increase current horizontal # part count by 1 hzntl += 1 i += 1 else: res += Y[j] * hzntl # increase current vertical # part count by 1 vert += 1 j += 1 # loop for horizontal array, if remains total = 0 while (i < m): total += X[i] i += 1 res += total * vert #loop for vertical array, if remains total = 0 while (j < n): total += Y[j] j += 1 res += total * hzntl return res # Driver program m = 6; n = 4X = [2, 1, 3, 1, 4] Y = [4, 1, 2] print(minimumCostOfBreaking(X, Y, m-1, n-1)) # This code is contributed by Anant Agarwal. |
C#
// C# program to divide a // board into m*n squares using System; class GFG { // method returns minimum cost to break board into // m*n squares static int minimumCostOfBreaking(int[] X, int[] Y, int m, int n) { int res = 0; // sort the horizontal cost in reverse order Array.Sort<int>(X, new Comparison<int>( (i1, i2) => i2.CompareTo(i1))); // sort the vertical cost in reverse order Array.Sort<int>(Y, new Comparison<int>( (i1, i2) => i2.CompareTo(i1))); // initialize current width as 1 int hzntl = 1, vert = 1; // loop untill one or both // cost array are processed int i = 0, j = 0; while (i < m && j < n) { if (X[i] > Y[j]) { res += X[i] * vert; // increase current horizontal // part count by 1 hzntl++; i++; } else { res += Y[j] * hzntl; // increase current vertical // part count by 1 vert++; j++; } } // loop for horizontal array, // if remains int total = 0; while (i < m) total += X[i++]; res += total * vert; // loop for vertical array, // if remains total = 0; while (j < n) total += Y[j++]; res += total * hzntl; return res; } // Driver program public static void Main(String []arg) { int m = 6, n = 4; int []X = {2, 1, 3, 1, 4}; int []Y = {4, 1, 2}; Console.WriteLine(minimumCostOfBreaking(X, Y, m-1, n-1)); } } // This code is contributed by Princi Singh |
Output:
42
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