覆盖N * M网格中给定职位的最低成本

给定一个n * m的网格并在网格中绘制一些极点的位置，任务是找到最小的绘制所有极点的成本。从一排移到另一排不涉及任何成本，而移至相邻的一列会产生1卢比的成本。

例子：

Input: n = 2, m = 2, noOfPos = 2
pos[0] = 0, 0
pos[1] = 0, 1

Output: 1
The grid is of 2*2 size and there are two poles at {0, 0} and {0, 1}. 
So we will start at {0, 0} and paint the pole and then go to
the next column to paint the pole at {0, 1} position which will
cost 1 rupee to move one column. 

Input: n = 2, m = 2, noOfPos = 2
pos[0] = {0, 0}
pos[1] = {1, 0}
Output: 0
Both poles are in the same column. So, no need to move to another column.

方法：由于移动列是唯一的成本，因此，如果我们转到任何列，我们将绘制该列中的所有极点，然后继续前进。因此，基本上，答案将是两个最远的列之间的差。

以下是所需的实现：

C++

// C++ implementation of the above approach
  
#include 
#include 
  
using namespace std;
  
    // Function to find the cost to paint all poles
    void find(int n,int m,int p,int q[2][2])
    {
        // To store all the columns,create list
        list  z ;
        int i ;
          
        for(i = 0;i < p;i++)
            z.push_back(q[i][1]);
          
        // sort in ascending order
        z.sort();
          
        // z.back() gives max value
        // z.front() gives min value
        cout << z.back() - z.front() <

---

Java

// Java implementation of the above approach 
  
import java.util.*;
class solution
{
  
    // Function to find the cost to paint all poles 
    static void find(int n,int m,int p,int q[][]) 
    { 
        // To store all the columns,create list 
        Vector  z= new Vector() ; 
        int i ; 
          
        for(i = 0;i < p;i++) 
            z.add(q[i][1]); 
          
        // sort in ascending order 
        Collections.sort(z); 
          
        // z.back() gives max value 
        // z.front() gives min value 
        System.out.print(z.get(z.size()-1) - z.get(0) ) ; 
    } 
      
    // Driver code 
    public static void main(String args[])
    { 
        int n = 2; 
        int m = 2; 
        int p = 2; 
          
        int q[][] = {{0,0},{0,1}} ; 
          
        find(n, m, p, q); 
          
           
    } 
  
  
  
}
//contributed by Arnab Kundu

Python3

# Function to find the cost to paint all poles
import math as ma
  
def find(n, m, p, q):
  
    # To store all the columns
    z =[]
    for i in range(p):
        z.append(q[i][1])
  
    print(max(z)-min(z))
      
n, m, p = 2, 2, 2
q =[(0, 0), (0, 1)]
find(n, m, p, q)

C#

// C# implementation of the above approach 
using System;
using System.Collections.Generic;
  
class GFG
{
  
    // Function to find the cost to paint all poles 
    static void find(int n, int m, int p, int [,]q) 
    { 
        // To store all the columns,create list 
        List  z = new List(); 
        int i; 
          
        for(i = 0; i < p; i++) 
            z.Add(q[i, 1]); 
          
        // sort in ascending order 
        z.Sort(); 
          
        // z.back() gives max value 
        // z.front() gives min value 
        Console.Write(z[z.Count-1] - z[0]); 
    } 
      
    // Driver code 
    public static void Main(String []args)
    { 
        int n = 2; 
        int m = 2; 
        int p = 2; 
          
        int [,]q = {{0, 0}, {0, 1}}; 
          
        find(n, m, p, q); 
    } 
}
  
// This code is contributed by PrinciRaj1992

输出：

1