矩阵中与给定两点等距的单元数
给定一个 N 行 M 列的矩阵，给定矩阵上的两点；任务是计算与给定两点等距的像元数量。水平方向或垂直方向或两个方向的任何遍历都被认为是有效的，但对角线路径无效。 举例:
输入:5 5 2 4 5 3 T5】输出: 5 说明: 在所有单元格中，这些是点(3，1)；(3, 2);(3, 3);(4, 4);(4，5) 满足给定条件。 输入: 4 3 2 3 4 1 输出: 4 解释: 在所有单元格中，这些是点(1，1)；(2, 1);(3, 2);(4，3) 满足给定条件。
接近 :
- 矩阵的每个单元都被遍历。
- 假设“A”是当前单元格和第一个点之间的距离，同样“B”是当前单元格和第二个点之间的距离。
- 使用曼哈顿距离计算两点之间的距离。如果 A & B 相等，则计数递增。
以下是上述方法的实现:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
int numberOfPoints(int N, int M, int x1,
int y1, int x2, int y2)
{
// Initializing count
int count = 0;
// Traversing through rows.
for (int i = 1; i <= N; i++) {
// Traversing through columns.
for (int j = 1; j <= M; j++) {
// By using Manhattan Distance, the distance between
// the current point to given two points is calculated.
// If distances are equal
// the count is incremented by 1.
if (abs(i - x1) + abs(j - y1)
== abs(i - x2) + abs(j - y2))
count++;
}
}
return count;
}
// Driver Code
int main()
{
int n = 5;
int m = 5;
int x1 = 2;
int y1 = 4;
int x2 = 5;
int y2 = 3;
cout << numberOfPoints(n, m, x1, y1, x2, y2);
}
Java 语言(一种计算机语言，尤用于创建网站)
//Java implementation of the above approach
import java.util.*;
import java.lang.Math;
public class GFG {
public static int numberOfPoints(int N, int M, int x1,
int y1, int x2, int y2)
{
int count = 0, i, j;
// Traversing through rows.
for (i = 1; i <= N; i++) {
// Traversing through columns.
for (j = 1; j <= M; j++) {
// By using Manhattan Distance, distance between
// current point to given two points is calculated
// If distances are equal
// the count is incremented by 1.
if (Math.abs(i - x1) + Math.abs(j - y1)
== Math.abs(i - x2) + Math.abs(j - y2))
count += 1;
}
}
return count;
}
// Driver Code
public static void main(String[] args)
{
int n = 5;
int m = 5;
int x1 = 2;
int y1 = 4;
int x2 = 5;
int y2 = 3;
System.out.println(numberOfPoints(n, m, x1, y1, x2, y2));
}
}
计算机编程语言
# Python implementation of the above approach
def numberPoints(N, M, x1, y1, x2, y2):
# Initializing count = 0
count = 0
# Traversing through rows.
for i in range(1, N + 1):
# Traversing through columns.
for j in range(1, M + 1):
# By using Manhattan Distance,
# distance between current point to
# given two points is calculated
# If distances are equal the
# count is incremented by 1.
if (abs(i - x1)+abs(j - y1)) == (abs(i - x2)+abs(j - y2)):
count += 1
return count
# Driver Code
N = 5
M = 5
x1 = 2
y1 = 4
x2 = 5
y2 = 3
print(numberPoints(N, M, x1, y1, x2, y2))
C
// C# implementation of the above approach
using System;
class GFG
{
static int numberOfPoints(int N, int M, int x1,
int y1, int x2, int y2)
{
int count = 0, i, j;
// Traversing through rows.
for (i = 1; i <= N; i++)
{
// Traversing through columns.
for (j = 1; j <= M; j++)
{
// By using Manhattan Distance, distance between
// current point to given two points is calculated
// If distances are equal
// the count is incremented by 1.
if (Math.Abs(i - x1) + Math.Abs(j - y1)
== Math.Abs(i - x2) + Math.Abs(j - y2))
count += 1;
}
}
return count;
}
// Driver Code
public static void Main()
{
int n = 5;
int m = 5;
int x1 = 2;
int y1 = 4;
int x2 = 5;
int y2 = 3;
Console.WriteLine(numberOfPoints(n, m, x1, y1, x2, y2));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// javascript implementation of the above approach
function numberOfPoints(N , M , x1 , y1 , x2 , y2)
{
var count = 0, i, j;
// Traversing through rows.
for (i = 1; i <= N; i++) {
// Traversing through columns.
for (j = 1; j <= M; j++) {
// By using Manhattan Distance, distance between
// current point to given two points is calculated
// If distances are equal
// the count is incremented by 1.
if (Math.abs(i - x1) + Math.abs(j - y1) == Math.abs(i - x2) + Math.abs(j - y2))
count += 1;
}
}
return count;
}
// Driver Code
var n = 5;
var m = 5;
var x1 = 2;
var y1 = 4;
var x2 = 5;
var y2 = 3;
document.write(numberOfPoints(n, m, x1, y1, x2, y2));
// This code is contributed by Rajput-Ji
</script>
Output:
5
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