C :

#include<stdio.h>
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int s,n;
		scanf("%d",&n);
		int a[2*n];
		for(int i=0;i<2*n;i++)
		{
			scanf("%d",&a[i]);
		}
		//对数组进行排序 
		for(int i=0;i<2*n;i++)
		{
			for(int j=i-1;j>=0;j--)
			{
				if(a[j]<=a[j+1])
				break;
				
				int temp=a[j+1];
				a[j+1]=a[j];
				a[j]=temp;
			}
		}
		//任意排序算法皆可，上述代码为其中一种排序示例
		//也可使用C++中的sort函数进行排序 
		s=(a[n-1]-a[0])+(a[2*n-1]-a[n]);
		printf("%d\n",s);
	}
	return 0;
}

C++ :

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e6+10;
typedef pair<int,int> PII;
typedef priority_queue<int , vector<int>, greater<int>> minqueue;  //从小到大 queue
typedef priority_queue<int, vector<int>, less<int>> maxqueue;      //从大到小 queue

int gcd(int a, int b){  //最大公因数 
    return b ? gcd(b, a % b) : a;
}

int lcm(int a, int b){ //最小公倍数 
    return a * b / __gcd(a, b);
}

int qmi(int base, int power, int p)  //快速幂求余 
{
	int result = 1;   
	while (power > 0)           
	{
		if (power & 1)         							
			result = result * base % p;   
		base = base * base % p ;       						
		power >>= 1;         						
	}
	return result % p;       
}

void solve()
{
	vector<int> vec;
	int n;
	cin>>n;
	int x;
	for(int i=0;i<2*n;i++){
		cin>>x;
		vec.push_back(x);
	}
	sort(vec.begin(),vec.end());
	int ans=vec[n-1]-vec[0]+vec[2*n-1]-vec[n];
	cout<<ans<<'\n';
}

signed main()
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int t;
	t=1;
	cin>>t; 
	while(t--)     
	{
		solve();
	}
	return 0;
}

Python :

# coding=utf-8
import heapq

def manhattan_distance(p1, p2):
    return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1])

def min_path_length(n, points):
    # 构建最小生成树
    # 采用 Prim 算法
    mst_cost = 0
    visited = [False] * n
    min_heap = [(0, 0)]  # (成本, 点索引)
    
    while min_heap:
        cost, u = heapq.heappop(min_heap)
        
        if visited[u]:
            continue
        visited[u] = True
        mst_cost += cost
        
        # 更新与当前点的连接边
        for v in range(n):
            if not visited[v]:
                distance = manhattan_distance(points[u], points[v])
                heapq.heappush(min_heap, (distance, v))
    
    return mst_cost

def solve():
    t = int(input())  # 测试用例数量
    for _ in range(t):
        n = int(input())  # 点的数量
        a = list(map(int, input().split()))  # 序列a
        
        # 将2n个数按顺序分为点的坐标 (x, y)
        a.sort()
        points = [(a[i], a[i + n]) for i in range(n)]
        
        # 计算最小路径长度
        print(min_path_length(n, points))

if __name__ == "__main__":
    solve()