最大化通过从给定数组中选择边形成的四边形的周长
给定一个大小为N的数组arr ，其中每个元素表示边的长度，任务是找到可以使用给定数组中的边创建的最大周长四边形。如果不能形成四边形，则打印-1。
例子：
Input: arr[ ] = {3, 1, 2, 4, 2, 1}
Output: 11
Explanation: A quadrilateral of perimeter 4 + 3 + 2 + 2 = 11 can be formed which is the maximum possible.
Input: arr[ ] = {1, 1, 2, 2, 20}
Output: 6
Input: arr[ ] = {1, 2, 4, 10}
Output: -1
方法：给定问题可以通过迭代给定数组中所有可能的边（a，b，c，d）组合来解决。可以看出，只有当三个较小边之和大于或等于最大边，或者四边形的周长大于或等于（2 * 最大的一面）。因此，遍历给定数组arr[]中(a, b, c, d)的所有可能值，并检查它是否形成有效的四边形。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find largest perimeter
// of a possible quadrilateral
int largestPerimeterQuad(int* arr, int n)
{
// Stores the final answer
int ans = -1;
// Loop to iterate over all possible
// sides (a, b, c, d) that can be
// formed from the given array
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
for (int k = j + 1; k < n; k++) {
for (int l = k + 1; l < n; l++) {
// If a quadrilateral can
// be formed from current
// selected sides
if (arr[i] + arr[j]
+ arr[k] + arr[l]
>= 2
* max(arr[i],
max(arr[j],
max(arr[k],
arr[l]))))
// Update maximum
ans = max(ans, arr[i] + arr[j]
+ arr[k]
+ arr[l]);
}
}
}
}
// Return Answer
return ans;
}
// Driver Code
int main()
{
int arr[] = { 3, 1, 2, 4, 2, 1 };
int n = sizeof(arr) / sizeof(int);
cout << largestPerimeterQuad(arr, n);
return 0;
} Java
// Java program for the above approach
import java.util.*;
public class GFG
{
// Function to find largest perimeter
// of a possible quadrilateral
static int largestPerimeterQuad(int []arr, int n)
{
// Stores the final answer
int ans = -1;
// Loop to iterate over all possible
// sides (a, b, c, d) that can be
// formed from the given array
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
for (int k = j + 1; k < n; k++) {
for (int l = k + 1; l < n; l++) {
// If a quadrilateral can
// be formed from current
// selected sides
if (arr[i] + arr[j]
+ arr[k] + arr[l]
>= 2
* Math.max(arr[i],
Math.max(arr[j],
Math.max(arr[k],
arr[l]))))
// Update maximum
ans = Math.max(ans, arr[i] + arr[j]
+ arr[k]
+ arr[l]);
}
}
}
}
// Return Answer
return ans;
}
// Driver Code
public static void main(String args[])
{
int []arr = { 3, 1, 2, 4, 2, 1 };
int n = arr.length;
System.out.print(largestPerimeterQuad(arr, n));
}
}
// This code is contributed by Samim Hossain Mondal.Python3
# Python program for the above approach
# Function to find largest perimeter
# of a possible quadrilateral
def largestPerimeterQuad(arr, n):
# Stores the final answer
ans = -1
# Loop to iterate over all possible
# sides (a, b, c, d) that can be
# formed from the given array
for i in range(n):
for j in range(i + 1, n):
for k in range(j + 1, n):
for l in range(k + 1, n):
# If a quadrilateral can
# be formed from current
# selected sides
if (arr[i] + arr[j]
+ arr[k] + arr[l]
>= 2 * max(arr[i],
max(arr[j],
max(arr[k],
arr[l])))):
# Update maximum
ans = max(ans, arr[i] + arr[j] + arr[k] + arr[l])
# Return Answer
return ans
# Driver Code
arr = [3, 1, 2, 4, 2, 1]
n = len(arr)
print(largestPerimeterQuad(arr, n))
# This code is contributed by gfgkingC#
// C# program for the above approach
using System;
class GFG
{
// Function to find largest perimeter
// of a possible quadrilateral
static int largestPerimeterQuad(int []arr, int n)
{
// Stores the final answer
int ans = -1;
// Loop to iterate over all possible
// sides (a, b, c, d) that can be
// formed from the given array
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
for (int k = j + 1; k < n; k++) {
for (int l = k + 1; l < n; l++) {
// If a quadrilateral can
// be formed from current
// selected sides
if (arr[i] + arr[j]
+ arr[k] + arr[l]
>= 2
* Math.Max(arr[i],
Math.Max(arr[j],
Math.Max(arr[k],
arr[l]))))
// Update maximum
ans = Math.Max(ans, arr[i] + arr[j]
+ arr[k]
+ arr[l]);
}
}
}
}
// Return Answer
return ans;
}
// Driver Code
public static void Main()
{
int []arr = { 3, 1, 2, 4, 2, 1 };
int n = arr.Length;
Console.Write(largestPerimeterQuad(arr, n));
}
}
// This code is contributed by Samim Hossain Mondal.Javascript
输出
11时间复杂度： O(N 4 )
辅助空间： O(1)