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
7 1 2 1 1 4 4
Output
1.0 4.0 5.0 3.5 4.5 5.0 5.0
Input
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：到达某一个节点，下一个访问的子代概率相同，按照dfs的顺序，求每个节点被访问到的时间
题解：对于对某个节点，设其父节点有另一分支a ，则该节点要么在a之前被访问，要么在a被全部访问完后，才被访问。同样我们把其父节点其他所有分支看成一个整体，该节点要么在这个整体之前要么在他之后，所以可得
该节点的期望就为： 其他分支的节点和 / 2 + 1 + 父节点的期望
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
const int N=1e5+10;
vector<int> v[N];
int n,son[N];
double dp[N];
void dfs1(int u)
{
son[u]=1;
for(int i=0;i<v[u].size();i++)
{
int to=v[u][i];
dfs1(to);
son[u]+=son[to];
}
}
void dfs2(int u)
{
for(int i=0;i<v[u].size();i++)
{
int to=v[u][i];
dp[to]=dp[u]+(double)(son[u]-son[to]-1)/2+1;
dfs2(to);
}
}
int main()
{
int fa;
scanf("%d",&n);
for(int i=2;i<=n;i++)
{
scanf("%d",&fa);
v[fa].pb(i);
}
dfs1(1);
dp[1]=1;
dfs2(1);
for(int i=1;i<=n;i++)printf("%.10f%c",dp[i]," \n"[i==n]);
return 0;
}