D. Puzzles
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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'snot 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 forin 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 arrayof 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
7121144
Output
1.04.05.03.54.55.05.0
Input
12112244331108
Output
1.05.05.56.57.58.08.07.07.56.57.58.0
#include <iostream>#include <cstdio>#include <set>//multiset的定義在set頭文件中usingnamespacestd;
int data[100005][2];//[0]自身以及子節點的數量[1]父節點double ans[100005];
struct Node {
int x;//父節點int y;//子節點
Node(int x, int y): x(x), y(y) {
}
intoperator < (Node const & n) const {//倒序return x > n.x;
}
};
multiset<Node> ms;
int main(int argc, charconst *argv[])
{
int n, temp;
scanf("%d", &n);
data[1][0] = 1;
for (int i = 2; i <= n; i++) {
scanf("%d", &temp);
data[i][0] = 1;
data[i][1] = temp;
Node n(temp, i);
ms.insert(n);
}
// 從樹的最底開始上升int cou = 0;
for (multiset<Node>::iterator i = ms.begin(); i != ms.end(); ++i) {
data[i->x][0] += data[i->y][0];
}
ans[1] = 1.0;
printf("1.0");
for (int i = 2; i <= n; i++) {
/**
* 第i個節點的期望 == 它的父節點的期望 + 在它自身的耗時(1) + 第i個節點的兄弟節點及它拓展的子節點耗時的期望(即兄弟節點及它拓展的子節點數 * 0.5)
* (兄弟節點及它拓展的子節點數 == i 的父節點 - i 節點拓展的樹的所有節點(包括i) - 1(父節點自身)),
* 這些節點每一個要麼在i之前被遍歷,要麼在i之後,於是期望值乘以0.5
*/
ans[i] = ans[data[i][1]] + 1 + (data[data[i][1]][0] - 1 - data[i][0]) * 0.5;
printf(" %.1f", ans[i]);
}
return0;
}