传送门

思路： 按照z排序，最后再按顺序计算(欧几里得)距离并相加即可。

代码实现：

#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9 + 7;
const int  N = 2e5 + 5;

int n;
double ans;

struct node{
	int x,y,z;
}a[N];

int cmp(node x, node y){
	return x.z < y.z;
}

double dist(node a, node b){
    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) + (a.z-b.z)*(a.z-b.z));
}

signed main()
{
	cin >> n;
	for(int i = 1;i <= n; i ++)  cin >> a[i].x >> a[i].y >> a[i].z;
	sort(a+1,a+n+1,cmp);
	for(int i = 2; i <= n; i ++) ans += dist(a[i-1], a[i]);
	printf("%.3lf",ans);

    return 0;
}