Problem:
Score : 300 points

Problem Statement

There are N towns in a coordinate plane. Town i is located at coordinates (x_i, y_i). The distance between Town i and Town j is \sqrt{\left(x_i-x_j\right)^2+\left(y_i-y_j\right)^2}.

There are N! possible paths to visit all of these towns once. Let the length of a path be the distance covered when we start at the first town in the path, visit the second, third, \dots, towns, and arrive at the last town (assume that we travel in a straight line from a town to another). Compute the average length of these N! paths.

Constraints

2 \leq N \leq 8

-1000 \leq x_i \leq 1000

-1000 \leq y_i \leq 1000

\left(x_i, y_i\right) \neq \left(x_j, y_j\right) (if i \neq j)

(Added 21:12 JST) All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the average length of the paths.
Your output will be judges as correct when the absolute difference from the judge's output is at most 10^{-6}.

Sample Input 1

3
0 0
1 0
0 1

Sample Output 1

2.2761423749

There are six paths to visit the towns: 1 → 2 → 3, 1 → 3 → 2, 2 → 1 → 3, 2 → 3 → 1, 3 → 1 → 2, and 3 → 2 → 1.

The length of the path 1 → 2 → 3 is \sqrt{\left(0-1\right)^2+\left(0-0\right)^2} + \sqrt{\left(1-0\right)^2+\left(0-1\right)^2} = 1+\sqrt{2}.

By calculating the lengths of the other paths in this way, we see that the average length of all routes is:

\frac{\left(1+\sqrt{2}\right)+\left(1+\sqrt{2}\right)+\left(2\right)+\left(1+\sqrt{2}\right)+\left(2\right)+\left(1+\sqrt{2}\right)}{6} = 2.276142...

Sample Input 2

2
-879 981
-866 890

Sample Output 2

91.9238815543

There are two paths to visit the towns: 1 → 2 and 2 → 1. These paths have the same length.

Sample Input 3

8
-406 10
512 859
494 362
-955 -475
128 553
-986 -885
763 77
449 310

Sample Output 3

7641.9817824387

Solution in Rust (1.15.1) (judge verdict: Accepted)

//! ----------------------------------------------
//! Framework <https://github.com/vain0x/procon>
//!
//! See the bottom of file for solution.
//! ----------------------------------------------

#![allow(unused_imports)]
#![allow(non_snake_case)]

use std::cell::RefCell;
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Display, Formatter, Write as FmtWrite};
use std::io::{stderr, stdin, BufRead, Write};
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;

/// Print values to standard error if debug mode.
#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), stderr());
            writeln!(err, "\x1B[33m{}\x1B[0m = {:?}", e, $e).unwrap()
        })*
    };
}

/// Read from standard input and parse each word.
/// - `read!(T, U, ..)` parses a line as a tuple of words.
/// - `read![[T]]` parses a line as an array of words.
/// - `read![..; N]` parses `N` lines, using `read!(..)` repeatedly.
#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

/// Read a line from standard input.
#[allow(dead_code)]
fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();

    #[allow(deprecated)]
    buf.trim_right().to_owned()
}

// -----------------------------------------------
// Solution
// -----------------------------------------------

type Point = (i64, i64);

fn square(x: i64) -> i64 {
    x * x
}

fn distance(p: Point, q: Point) -> f64 {
    let (px, py) = p;
    let (qx, qy) = q;

    ((square(px - qx) + square(py - qy)) as f64).sqrt()
}

struct Solver {
    /// 街の個数
    N: usize,
    /// 街の座標のリスト
    points: Vec<Point>,

    //.AsMut perm に含まれる街の数
    k: usize,

    /// perm[0..k] は街の順列の一部。
    /// 最初に perm[0] に訪れて、
    /// そこから perm[1], perm[2], ... の順番で移動する、
    /// ということが決定事項であることを表す。
    perm: Vec<usize>,

    /// 街 x がすでに perm に含まれていたら done[x] = true
    done: Vec<bool>,

    /// 移動距離の総和
    total: f64,
    /// 経路数 (最終的に N! になる)
    count: i64,
}

impl Solver {
    fn distance(&self) -> f64 {
        let mut sum = 0.0;
        for w in self.perm.windows(2) {
            sum += distance(self.points[w[0]], self.points[w[1]]);
        }
        sum
    }

    // perm[0..i] が街の順列の一部になっていて、
    // 残りの perm[i..N]
    fn dfs(&mut self) {
        let N = self.N;

        // 移動経路が1つに確定
        if self.k == N {
            self.total += self.distance();
            self.count += 1;
            return;
        }

        // 次に移動する街を全通り試す。
        for i in 0..N {
            // 移動済みの街は訪れない
            if self.done[i] {
                continue;
            }

            self.perm[self.k] = i;
            self.done[i] = true;
            self.k += 1;

            // 残りの移動順を決めて移動経路の総和を増やす。
            self.dfs();

            // 状態を戻す。
            self.k -= 1;
            self.done[i] = false;
        }
    }

    fn solve(&mut self) -> f64 {
        self.dfs();

        // 平均
        self.total / self.count as f64
    }
}

fn main() {
    let N = read!(usize);
    let points = read![i64, i64; N];

    let average = Solver {
        N: N,
        points: points,
        k: 0,
        perm: vec![0; N],
        done: vec![false; N],
        total: 0.0,
        count: 0,
    }
    .solve();

    println!("{}", average)
}