从满足给定条件的数组中计算坐标数
给定一个由笛卡尔平面中的 N 坐标组成的数组 arr[] ，任务是找到满足以下所有条件的坐标数，如 (X，Y) :
- 所有可能的 arr[i][0] 必须小于 X ， arr[i][1] 必须等于 Y 。
- 所有可能的 arr[i][0] 必须大于 X ， arr[i][1] 必须等于 Y 。
- 所有可能的 arr[i][0] 必须小于 Y ， arr[i][1] 必须等于 X 。
- 所有可能的 arr[i][0] 必须大于 Y ， arr[i][1] 必须等于 X 。
示例:
输入: arr[][] = {{0，0}，{0，1}，{1，0}，{0，-1}，{-1，0}} 输出: 1 说明:基于以下条件，只存在一个满足给定条件的坐标，即 (0，0) :
- arr[2] = {1，0}:由于 arr[2]0 > X(= 0)和 arr[2]1= Y(= 0)，因此满足条件 1。
- arr[4] = {-1，0}:由于 arr[4]0 < 0 且 arr[4]1 == Y(= 0)，因此满足条件 2。
- arr[1] = {0，1}:由于 arr[1]0 == X(= 0)和 arr[1]1 > Y(= 0)，因此满足条件 3。
- arr[3] = {0，-1}:由于 arr[3]0 == X(= 0)和 arr[3]1 < Y(= 0)，因此满足条件 4。
因此，总点数为 1。
输入: arr[][] = {{1，0}，{2，0}，{1，1}，{1，-1}} 输出: 0
逼近:给定问题可以通过考虑每一个坐标来求解，比如说 (arr[i][0]、arr[i][1]) 作为合成坐标，如果坐标 (arr[i][0]、arr[i][1]) 满足所有给定条件，那么就统计当前坐标。检查所有坐标后，打印获得的总计数值。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number of
// coordinates from a given set
// that satisfies the given conditions
int centralPoints(int arr[][2], int N)
{
// Stores the count of central points
int count = 0;
// Store the count of
// each x and y coordinates
int c1, c2, c3, c4;
// Find all possible pairs
for (int i = 0; i < N; i++) {
// Initialize variables c1, c2,
// c3, c4 to define the status
// of conditions
c1 = 0, c2 = 0, c3 = 0;
c4 = 0;
// Stores value of each point
int x = arr[i][0];
int y = arr[i][1];
// Check the conditions for
// each point by generating
// all possible pairs
for (int j = 0; j < N; j++) {
// If arr[j][0] > x and
// arr[j][1] == y
if (arr[j][0] > x
&& arr[j][1] == y) {
c1 = 1;
}
// If arr[j][0] < x and
// arr[j][1] == y
if (arr[j][1] > y
&& arr[j][0] == x) {
c2 = 1;
}
// If arr[j][1] > y and
// arr[j][0] == x
if (arr[j][0] < x
&& arr[j][1] == y) {
c3 = 1;
}
// If arr[j][1] < y and
// arr[j][0] == x
if (arr[j][1] < y
&& arr[j][0] == x) {
c4 = 1;
}
}
// If all conditions satisfy
// then point is central point
if (c1 + c2 + c3 + c4 == 4) {
// Increment the count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
int main()
{
int arr[4][2] = { { 1, 0 }, { 2, 0 },
{ 1, 1 }, { 1, -1 } };
int N = sizeof(arr) / sizeof(arr[0]);
cout << centralPoints(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Function to count the number of
// coordinates from a given set
// that satisfies the given conditions
static int centralPoints(int arr[][], int N)
{
// Stores the count of central points
int count = 0;
// Store the count of
// each x and y coordinates
int c1, c2, c3, c4;
// Find all possible pairs
for (int i = 0; i < N; i++) {
// Initialize variables c1, c2,
// c3, c4 to define the status
// of conditions
c1 = 0;
c2 = 0;
c3 = 0;
c4 = 0;
// Stores value of each point
int x = arr[i][0];
int y = arr[i][1];
// Check the conditions for
// each point by generating
// all possible pairs
for (int j = 0; j < N; j++) {
// If arr[j][0] > x and
// arr[j][1] == y
if (arr[j][0] > x && arr[j][1] == y) {
c1 = 1;
}
// If arr[j][0] < x and
// arr[j][1] == y
if (arr[j][1] > y && arr[j][0] == x) {
c2 = 1;
}
// If arr[j][1] > y and
// arr[j][0] == x
if (arr[j][0] < x && arr[j][1] == y) {
c3 = 1;
}
// If arr[j][1] < y and
// arr[j][0] == x
if (arr[j][1] < y && arr[j][0] == x) {
c4 = 1;
}
}
// If all conditions satisfy
// then point is central point
if (c1 + c2 + c3 + c4 == 4) {
// Increment the count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
public static void main(String[] args)
{
int arr[][]
= { { 1, 0 }, { 2, 0 }, { 1, 1 }, { 1, -1 } };
int N = arr.length;
System.out.println(centralPoints(arr, N));
}
}
// This code is contributed by Kingash.
Python 3
# Python3 program for the above approach
# Function to count the number of
# coordinates from a given set
# that satisfies the given conditions
def centralPoints(arr, N):
# Stores the count of central points
count = 0
# Find all possible pairs
for i in range(N):
# Initialize variables c1, c2,
# c3, c4 to define the status
# of conditions
c1 = 0
c2 = 0
c3 = 0
c4 = 0
# Stores value of each point
x = arr[i][0]
y = arr[i][1]
# Check the conditions for
# each point by generating
# all possible pairs
for j in range(N):
# If arr[j][0] > x and
# arr[j][1] == y
if (arr[j][0] > x and arr[j][1] == y):
c1 = 1
# If arr[j][0] < x and
# arr[j][1] == y
if (arr[j][1] > y and arr[j][0] == x):
c2 = 1
# If arr[j][1] > y and
# arr[j][0] == x
if (arr[j][0] < x and arr[j][1] == y):
c3 = 1
# If arr[j][1] < y and
# arr[j][0] == x
if (arr[j][1] < y and arr[j][0] == x):
c4 = 1
# If all conditions satisfy
# then point is central point
if (c1 + c2 + c3 + c4 == 4):
# Increment the count by 1
count += 1
# Return the count
return count
# Driver Code
if __name__ == '__main__':
arr = [ [ 1, 0 ], [ 2, 0 ],
[ 1, 1 ], [ 1, -1 ] ]
N = len(arr)
print(centralPoints(arr, N));
# This code is contributed by SURENDRA_GANGWAR
C
// C# program for the above approach
using System;
class GFG{
// Function to count the number of
// coordinates from a given set
// that satisfies the given conditions
static int centralPoints(int[,] arr, int N)
{
// Stores the count of central points
int count = 0;
// Store the count of
// each x and y coordinates
int c1, c2, c3, c4;
// Find all possible pairs
for(int i = 0; i < N; i++)
{
// Initialize variables c1, c2,
// c3, c4 to define the status
// of conditions
c1 = 0;
c2 = 0;
c3 = 0;
c4 = 0;
// Stores value of each point
int x = arr[i, 0];
int y = arr[i, 1];
// Check the conditions for
// each point by generating
// all possible pairs
for(int j = 0; j < N; j++)
{
// If arr[j][0] > x and
// arr[j][1] == y
if (arr[j, 0] > x && arr[j, 1] == y)
{
c1 = 1;
}
// If arr[j][0] < x and
// arr[j][1] == y
if (arr[j, 1] > y && arr[j, 0] == x)
{
c2 = 1;
}
// If arr[j][1] > y and
// arr[j][0] == x
if (arr[j, 0] < x && arr[j, 1] == y)
{
c3 = 1;
}
// If arr[j][1] < y and
// arr[j][0] == x
if (arr[j, 1] < y && arr[j, 0] == x)
{
c4 = 1;
}
}
// If all conditions satisfy
// then point is central point
if (c1 + c2 + c3 + c4 == 4)
{
// Increment the count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
public static void Main(string[] args)
{
int[,] arr = { { 1, 0 }, { 2, 0 },
{ 1, 1 }, { 1, -1 } };
int N = arr.GetLength(0);
Console.WriteLine(centralPoints(arr, N));
}
}
// This code is contributed by ukasp
java 描述语言
<script>
// Javascript program to implement the
// above approach
// Function to count the number of
// coordinates from a given set
// that satisfies the given conditions
function centralPoints(arr, N)
{
// Stores the count of central points
let count = 0;
// Store the count of
// each x and y coordinates
let c1, c2, c3, c4;
// Find all possible pairs
for (let i = 0; i < N; i++) {
// Initialize variables c1, c2,
// c3, c4 to define the status
// of conditions
c1 = 0;
c2 = 0;
c3 = 0;
c4 = 0;
// Stores value of each point
let x = arr[i][0];
let y = arr[i][1];
// Check the conditions for
// each point by generating
// all possible pairs
for (let j = 0; j < N; j++) {
// If arr[j][0] > x and
// arr[j][1] == y
if (arr[j][0] > x && arr[j][1] == y) {
c1 = 1;
}
// If arr[j][0] < x and
// arr[j][1] == y
if (arr[j][1] > y && arr[j][0] == x) {
c2 = 1;
}
// If arr[j][1] > y and
// arr[j][0] == x
if (arr[j][0] < x && arr[j][1] == y) {
c3 = 1;
}
// If arr[j][1] < y and
// arr[j][0] == x
if (arr[j][1] < y && arr[j][0] == x) {
c4 = 1;
}
}
// If all conditions satisfy
// then point is central point
if (c1 + c2 + c3 + c4 == 4) {
// Increment the count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
let arr = [[ 1, 0 ], [ 2, 0 ], [ 1, 1 ], [ 1, -1 ]];
let N = arr.length;
document.write(centralPoints(arr, N));
// This code is contributed by splevel62.
</script>
Output:
0
时间复杂度:O(N2) 辅助空间: O(1)
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