D. Puzzles [排列组合]

Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 roads between them. Cities and roads of USC form a rooted tree (Barney's not sure why it is rooted). Root of the tree is the city number 1. Thus if one will start his journey from city 1, he can visit any city he wants by following roads.

Some girl has stolen Barney's heart, and Barney wants to find her. He starts looking for in the root of the tree and (since he is Barney Stinson not a random guy), he uses a random DFS to search in the cities. A pseudo code of this algorithm is as follows:

let starting_time be an array of length n
current_time = 0
dfs(v):
	current_time = current_time + 1
	starting_time[v] = current_time
	shuffle children[v] randomly (each permutation with equal possibility)
	// children[v] is vector of children cities of city v
	for u in children[v]:
		dfs(u)

As told before, Barney will start his journey in the root of the tree (equivalent to call dfs(1)).

Now Barney needs to pack a backpack and so he wants to know more about his upcoming journey: for every city i, Barney wants to know the expected value of starting_time[i]. He's a friend of Jon Snow and knows nothing, that's why he asked for your help.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 105) — the number of cities in USC.

The second line contains n - 1 integers p2, p3, ..., pn (1 ≤ pi < i), where pi is the number of the parent city of city number i in the tree, meaning there is a road between cities numbered pi and i in USC.

Output

In the first and only line of output print n numbers, where i-th number is the expected value of starting_time[i].

Your answer for each city will be considered correct if its absolute or relative error does not exceed 10 - 6.

         Examples       
           input                       Copy          
7
1 2 1 1 4 4
           output         
1.0 4.0 5.0 3.5 4.5 5.0 5.0 
           input                       Copy          
12
1 1 2 2 4 4 3 3 1 10 8
           output         
1.0 5.0 5.5 6.5 7.5 8.0 8.0 7.0 7.5 6.5 7.5 8.0 
          
         

题意: 每到一个节点会DFS下去,每次花费的时间是1,问对于每个点花费的时间概率是多少

思路: 对于一个父亲的儿子们,有n!种遍历方式,但是根据全排列的概率均匀分布,每一个节点在兄弟节点后面的概率都是1/2 , 那么对于当点的概率 = ans[father]+1+sigma(兄弟节点的个数)/2

#include<bits/stdc++.h>
#define FAST ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N=1e5+5;
const int MOD=1e9+7;
const int INF=0x3f3f3f3f;

vector <int> edge[MAX_N];
int cnt[MAX_N];
double ans[MAX_N];

void dfs1(int u){///cul the all node under u
    cnt[u]=(int)edge[u].size();
    for(int i=0;i<(int)edge[u].size();++i){
        int v=edge[u][i];
        dfs1(v);
        cnt[u]+=cnt[v];
    }
}

void dfs2(int u){
    for(int i=0;i<(int)edge[u].size();++i){
        int v=edge[u][i];
        ans[v]=ans[u]+1+1.0*(cnt[u]-cnt[v]-1)/2;
        dfs2(v);
    }
}

int main(void){
    int n;
    cin >> n;
    for(int i=2;i<=n;i++){
        int v;
        scanf("%d",&v);
        edge[v].push_back(i);
    }
    dfs1(1);
    ans[1]=1;
    dfs2(1);
    for(int i=1;i<=n;i++)
        printf("%.8f ",ans[i]);
}