给定圆的圆周和数组pos [] ，该数组标记圆上N个点相对于固定点沿顺时针方向的距离。我们必须找到可以访问所有点的最小距离。我们可以从任何一点开始。
例子：
Input: circumference = 20, pos = [3, 6, 9]
Output: min path cost =6
Explanation:
If we start from 3, we go to 6 and then we go to 9. Threfore, total path cost is 3 units for first movement and 3 units for second movement which sums up to 6.
Input:circumference=20 pos = [3, 6, 19]
Output: min path cost = 7
Explanation :
If we start from 19 and we go to 3 it will cost 4 units because we go from 19 -> 20 -> 1 -> 2 -> 3 which gives 4 units, and then 3 to 6 which gives 3 units. In total minimum cost will be 4 + 3 = 7.
方法 ：
为了解决上述问题，我们必须遵循以下步骤：
- 对标记圆上N个点的距离的数组进行排序。
- 通过将N个元素的值加上arr [i + n] =周长+ arr [i]来使数组大小增加两次。
- 求出值i的所有有效迭代的最小值(arr [i +(n-1)] – arr [i]) 。
下面是上述方法的实现：
C++
// C++ implementation to find the
// Minimum Cost Path to visit all nodes
// situated at the Circumference of
// Circular Road
#include
using namespace std;
// Function to find the minimum cost
int minCost(int arr[], int n, int circumference)
{
// Sort the given array
sort(arr, arr + n);
// Initialise a new array of double size
int arr2[2 * n];
// Fill the array elements
for (int i = 0; i < n; i++) {
arr2[i] = arr[i];
arr2[i + n] = arr[i] + circumference;
}
// Find the minimum path cost
int res = INT_MAX;
for (int i = 0; i < n; i++)
res = min(res, arr2[i + (n - 1)] - arr2[i]);
// Return the final result
return res;
}
// Driver code
int main()
{
int arr[] = { 19, 3, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
int circumference = 20;
cout << minCost(arr, n, circumference);
return 0;
} Java
// Java implementation to find the
// Minimum Cost Path to visit all nodes
// situated at the Circumference of
// Circular Road
import java.util.*;
import java. util. Arrays;
class GFG {
// Function to find the minimum cost
static int minCost(int arr[], int n,
int circumference)
{
// Sort the given array
Arrays.sort(arr);
// Initialise a new array of double size
int[] arr2 = new int[2 * n];
// Fill the array elements
for(int i = 0; i < n; i++)
{
arr2[i] = arr[i];
arr2[i + n] = arr[i] + circumference;
}
// Find the minimum path cost
int res = Integer.MAX_VALUE;
for(int i = 0; i < n; i++)
res = Math.min(res,
arr2[i + (n - 1)] -
arr2[i]);
// Return the final result
return res;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 19, 3, 6 };
int n = arr.length;
int circumference = 20;
System.out.println(minCost(arr, n,
circumference));
}
}
// This code is contributed by ANKITKUMAR34Python3
# Python3 implementation to find the
# minimum cost path to visit all nodes
# situated at the circumference of
# circular Road
# Function to find the minimum cost
def minCost(arr, n, circumference):
# Sort the given array
arr.sort()
#Initialise a new array of double size
arr2 = [0] * (2 * n)
# Fill the array elements
for i in range(n):
arr2[i] = arr[i]
arr2[i + n] = arr[i] + circumference
# Find the minimum path cost
res = 9999999999999999999;
for i in range(n):
res = min(res,
arr2[i + (n - 1)] -
arr2[i]);
# Return the final result
return res;
# Driver code
arr = [ 19, 3, 6 ];
n = len(arr)
circumference = 20;
print(minCost(arr, n, circumference))
# This code is contributed by ANKITKUMAR34C#
// C# implementation to find the
// Minimum Cost Path to visit all nodes
// situated at the Circumference of
// Circular Road
using System;
class GFG{
// Function to find the minimum cost
static int minCost(int []arr, int n,
int circumference)
{
// Sort the given array
Array.Sort(arr);
// Initialise a new array of double size
int[] arr2 = new int[2 * n];
// Fill the array elements
for(int i = 0; i < n; i++)
{
arr2[i] = arr[i];
arr2[i + n] = arr[i] + circumference;
}
// Find the minimum path cost
int res = int.MaxValue;
for(int i = 0; i < n; i++)
res = Math.Min(res, arr2[i + (n - 1)] -
arr2[i]);
// Return the readonly result
return res;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 19, 3, 6 };
int n = arr.Length;
int circumference = 20;
Console.WriteLine(minCost(arr, n, circumference));
}
}
// This code is contributed by Princi Singh输出：
7