D. Puzzles
time limit per test
memory limit per test
input
output
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
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
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
d[i]表示以第i个节点为根的树上节点个数
starting_time[i] = ans[i] = ans[j] + (d[j] - d[i] - 1) * 0.5 + 1;(j是i的父节点)
#include <bits/stdc++.h>
#define maxn 100005
#define MOD 1000000007
using namespace std;
typedef long long ll;
int d[maxn];
double ans[maxn];
vector<int> v[maxn];
void dfs(int j, int f){
d[j] = 1;
for(int i = 0; i < v[j].size(); i++){
if(f == v[j][i])
continue;
dfs(v[j][i], j);
d[j] += d[v[j][i]];
}
}
void Dfs(int j, int f){
for(int i = 0; i < v[j].size(); i++){
if(f == v[j][i])
continue;
int p = v[j][i];
ans[p] = ans[j] + (d[j] - d[p] - 1) * 0.5 + 1;
Dfs(p, j);
}
}
int main(){
// freopen("in.txt", "r", stdin);
int n, a;
scanf("%d", &n);
for(int i = 2; i <= n; i++){
scanf("%d", &a);
v[a].push_back(i);
}
dfs(1, -1);
ans[1] = 1;
Dfs(1, -1);
printf("%.1lf", ans[1]);
for(int i = 2; i <= n; i++)
printf(" %.1lf", ans[i]);
puts("");
return 0;
}