给定一个N行M列的矩阵,给定矩阵上的两个点;任务是计算与给定两点等距的单元格数量。任何在水平方向或垂直方向或双向的遍历都被认为是有效的,但对角线路径是无效的。
例子:
Input: 5 5
2 4
5 3
Output: 5
Explanation:
Out of all cells, these are the points (3, 1);(3, 2);(3, 3);(4, 4);(4, 5)
which satisfy given condition.
Input: 4 3
2 3
4 1
Output: 4
Explanation:
Out of all cells, these are the points (1, 1);(2, 1);(3, 2);(4, 3)
which satisfy given condition.
方法:
- 遍历矩阵的每个单元格。
- 让’A’ 是当前单元格和第一个点之间的距离,类似地’B’ 是当前单元格和第二个点之间的距离。
- 两点之间的距离使用曼哈顿距离计算。如果 A 和 B 相等,则计数递增。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include
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
//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));
}
}
Python
# 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
Javascript
输出:
5
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