给定一个由N 个整数组成的数组arr[] ，每对点(x, y)将半圆的端点表示为(x, 0)和(y, 0) ，任务是检查是否有任何半圆对相交与否。如果发现是真的，则打印“是” 。否则，打印“否” 。
例子：
Input: arr[] = {0, 15, 5, 10}
Output: No
Input: arr[] = {0, 10, 5, 15}
Output: Yes
方法：可以通过检查任何对半圆是否相交，然后相应地打印元素来解决给定的问题。请按照以下步骤解决问题：
- 将所有可能的半圆及其对应的x 坐标和y 坐标存储在一个向量中。
- 现在，生成上述向量的所有可能对并执行以下步骤：
- 让这对坐标为X和Y，并使用以下条件检查圆是否相交：
- 如果(X[0] < Y[0] , X[1] < Y[1] and Y[0] < X[2])或(Y[0] < X[0] , Y[1 ] ] < X[1]和X[0] < Y[1]) ，那么半圆相交。否则，它不会相交。
- 如果以上两个圆圈相交，则打印“Yes”并跳出循环。
- 让这对坐标为X和Y，并使用以下条件检查圆是否相交：
- 完成上述步骤后，如果任何一对圆不相交，则打印“No” 。
C++
// C++ program for the above approach
#include
#include
using namespace std;
// Function to check if any pairs of
// the semicircles intersects or not
bool checkIntersection(int arr[], int N)
{
// Stores the coordinates of all
// the semicircles
vector > vec;
for (int i = 0; i < N - 1; i++) {
// x and y are coordinates
int x = arr[i], y = arr[i + 1];
// Store the minimum and maximum
// value of the pair
int minn = min(x, y);
int maxx = max(x, y);
// Push the pair in vector
vec.push_back({ minn, maxx });
}
// Compare one pair with other pairs
for (int i = 0; i < vec.size(); i++) {
pair x = vec[i];
// Generating the second pair
for (int j = 0;
j < vec.size(); j++) {
// Extract all the second
// pairs one by one
pair y = vec[j];
// 1st condition
bool cond1
= (x.first < y.first
and x.second < y.second
and y.first < x.second);
// 2nd condition
bool cond2
= (y.first < x.first
and y.second < x.second
and x.first < y.second);
// If any one condition
// is true
if (cond1 or cond2) {
return true;
}
}
}
// If any pair of semicircles
// doesn't exists
return false;
}
// Driver Code
int main()
{
int arr[] = { 0, 15, 5, 10 };
int N = sizeof(arr) / sizeof(int);
if (checkIntersection(arr, N))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to check if any pairs of
// the semicircles intersects or not
static boolean checkIntersection(int arr[], int N)
{
// Stores the coordinates of all
// the semicircles
Vector vec = new Vector<>();
for(int i = 0; i < N - 1; i++)
{
// x and y are coordinates
int x = arr[i], y = arr[i + 1];
// Store the minimum and maximum
// value of the pair
int minn = Math.min(x, y);
int maxx = Math.max(x, y);
// Push the pair in vector
vec.add(new pair( minn, maxx ));
}
// Compare one pair with other pairs
for(int i = 0; i < vec.size(); i++)
{
pair x = vec.elementAt(i);
// Generating the second pair
for(int j = 0;
j < vec.size(); j++)
{
// Extract all the second
// pairs one by one
pair y = vec.elementAt(j);
// 1st condition
boolean cond1 = (x.first < y.first &&
x.second < y.second &&
y.first < x.second);
// 2nd condition
boolean cond2 = (y.first < x.first &&
y.second < x.second &&
x.first < y.second);
// If any one condition
// is true
if (cond1 || cond2)
{
return true;
}
}
}
// If any pair of semicircles
// doesn't exists
return false;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 0, 15, 5, 10 };
int N = arr.length;
if (checkIntersection(arr, N))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Amit Katiyar Python3
# Python3 program for the above approach
# Function to check if any pairs of
# the semicircles intersects or not
def checkIntersection(arr, N):
# Stores the coordinates of all
# the semicircles
vec = []
for i in range(N - 1):
# x and y are coordinates
x = arr[i]
y = arr[i + 1]
# Store the minimum and maximum
# value of the pair
minn = min(x, y)
maxx = max(x, y)
# Push the pair in vector
vec.append([minn, maxx])
# Compare one pair with other pairs
for i in range(len(vec)):
x = vec[i]
# Generating the second pair
for j in range(len(vec)):
# Extract all the second
# pairs one by one
y = vec[j]
# 1st condition
cond1 = (x[0] < y[0] and
x[1] < y[1] and
y[0] < x[1])
# 2nd condition
cond2 = (y[0] < x[0] and
y[1] < x[1] and
x[0] < y[1])
# If any one condition
# is true
if (cond1 or cond2):
return True
# If any pair of semicircles
# doesn't exists
return False
# Driver Code
if __name__ == '__main__':
arr = [ 0, 15, 5, 10 ]
N = len(arr)
if (checkIntersection(arr, N)):
print("Yes")
else:
print("No")
# This code is contributed by ipg2016107C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to check if any pairs of
// the semicircles intersects or not
static bool checkIntersection(int []arr, int N)
{
// Stores the coordinates of all
// the semicircles
List vec = new List();
for(int i = 0; i < N - 1; i++)
{
// x and y are coordinates
int x = arr[i], y = arr[i + 1];
// Store the minimum and maximum
// value of the pair
int minn = Math.Min(x, y);
int maxx = Math.Max(x, y);
// Push the pair in vector
vec.Add(new pair( minn, maxx ));
}
// Compare one pair with other pairs
for(int i = 0; i < vec.Count; i++)
{
pair x = vec[i];
// Generating the second pair
for(int j = 0;
j < vec.Count; j++)
{
// Extract all the second
// pairs one by one
pair y = vec[j];
// 1st condition
bool cond1 = (x.first < y.first &&
x.second < y.second &&
y.first < x.second);
// 2nd condition
bool cond2 = (y.first < x.first &&
y.second < x.second &&
x.first < y.second);
// If any one condition
// is true
if (cond1 || cond2)
{
return true;
}
}
}
// If any pair of semicircles
// doesn't exists
return false;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 0, 15, 5, 10 };
int N = arr.Length;
if (checkIntersection(arr, N))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Princi Singh 输出：
No时间复杂度： O(N 2 )
辅助空间： O(N)
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