使用水平、垂直和对角线移动最小化到达 Matrix 中的单元格的成本

给定矩阵的两个点P ₁ (x ₁ , y ₁ )和P ₂ (x ₂ , y ₂ ) ，任务是找到从P ₁到达P ₂的最小成本：

任何方向的水平或垂直移动需要 1 个单位
任何方向的对角线运动消耗 0 个单位。

例子：

Input: P₁ = {7, 4}, P₂ = {4, 4}
Output: 1
Explanation: The movements are given below given below:

Movements are (7, 4) -> (6, 5) -> (5, 5) -> (4, 4).
As there is only 1 vertical movement so cost will be 1.

Input: P₁ = {1, 2}, P₂ = {2, 2}
Output: 1

方法：运动应该使得水平或垂直运动最小。这可以从以下观察中确定：

If the horizontal distance is h = (y₂ – y₁) and the vertical distance between the points is v = (x₂ – x₁):
If |h – v| is even, only diagonal movements are enough.
Because horizontal or vertical positions can be maintained with two opposite diagonal movements as shown in the image below:

maintaining vertical position

And when horizontal and vertical distances become same, then can directly move to P₂ using diagonal movements.
But in case this value is odd then one vertical or horizontal movement is required to make it even.

编程需要懂一点英语

为此，请执行以下步骤：

求从P ₁到P ₂的垂直距离。假设它是vert 。
求从P ₁到P ₂的水平距离。假设它是hori 。
如果|hori – vert| ，总有一条从源到目的地的路，通过对角线移动。甚至。
但是如果|hori – vert|是奇数，则需要水平或垂直 1 步之后，仅对角线移动就足够了。

下面是上述方法的实现。

C++

// C++ algorithm for above approach
#include 
using namespace std;
 
// Function to find Minimum Cost
int minCostToExit(int x1, int y1, int x2, int y2)
{
    // Finding vertical distance
    // from destination
    int vert = abs(x2 - x1);
 
    // Finding horizontal distance
    // from destination
    int hori = abs(y1 - y2);
 
    // Finding the minimum cost
    if (abs(hori - vert) % 2 == 0)
        return 0;
    else
        return 1;
}
 
// Driver code
int main()
{
    int x1 = 7;
    int y1 = 4;
    int x2 = 4, y2 = 4;
    int cost = minCostToExit(x1, y1, x2, y2);
    cout << cost;
    return 0;
}

Java

// Java algorithm for above approach
class GFG {
 
  // Function to find Minimum Cost
  static int minCostToExit(int x1, int y1,
                           int x2, int y2)
  {
 
    // Finding vertical distance
    // from destination
    int vert = Math.abs(x2 - x1);
 
    // Finding horizontal distance
    // from destination
    int hori = Math.abs(y1 - y2);
 
    // Finding the minimum cost
    if (Math.abs(hori - vert) % 2 == 0)
      return 0;
    else
      return 1;
  }
 
  // Driver code
  public static void main(String args[]) {
    int x1 = 7;
    int y1 = 4;
    int x2 = 4, y2 = 4;
    int cost = minCostToExit(x1, y1, x2, y2);
    System.out.println(cost);
  }
}
 
// This code is contributed by Saurabh Jaiswal

Python3

# Python algorithm for above approach
import math as Math
 
# Function to find Minimum Cost
def minCostToExit (x1, y1, x2, y2):
 
    # Finding vertical distance
    # from destination
    vert = Math.fabs(x2 - x1);
 
    # Finding horizontal distance
    # from destination
    hori = Math.fabs(y1 - y2);
 
    # Finding the minimum cost
    if (Math.fabs(hori - vert) % 2 == 0):
        return 0;
    else:
        return 1;
 
# Driver code
x1 = 7
y1 = 4
x2 = 4
y2 = 4;
cost = minCostToExit(x1, y1, x2, y2);
print(cost);
 
# This code is contributed by gfgking

C#

// C# algorithm for above approach
using System;
class GFG {
 
  // Function to find Minimum Cost
  static int minCostToExit(int x1, int y1, int x2, int y2)
  {
    // Finding vertical distance
    // from destination
    int vert = Math.Abs(x2 - x1);
 
    // Finding horizontal distance
    // from destination
    int hori = Math.Abs(y1 - y2);
 
    // Finding the minimum cost
    if (Math.Abs(hori - vert) % 2 == 0)
      return 0;
    else
      return 1;
  }
 
  // Driver code
  public static void Main()
  {
    int x1 = 7;
    int y1 = 4;
    int x2 = 4, y2 = 4;
    int cost = minCostToExit(x1, y1, x2, y2);
    Console.Write(cost);
  }
}
 
// This code is contributed by ukasp.

Javascript

输出

时间复杂度： O(1)
辅助空间： O(1)