【Codeforces Round 362 (Div 2)D】【樹的遍歷 概率均分思想】Puzzles 兄弟節點的等概率遍歷下 樹的遍歷每點期望時間戳

Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1through 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 calldfs(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.

7
1 2 1 1 4 4

1.0 4.0 5.0 3.5 4.5 5.0 5.0 

12
1 1 2 2 4 4 3 3 1 10 8

1.0 5.0 5.5 6.5 7.5 8.0 8.0 7.0 7.5 6.5 7.5 8.0

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 1e5+10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
vector<int>a[N];
double ans[N];
int sz[N];
void dfs1(int x)
{
	sz[x] = 0;
	for (int i = a[x].size() - 1; ~i; --i)
	{
		int y = a[x][i];
		dfs1(y);
		sz[x] += sz[y];
	}
	++sz[x];
}
void dfs2(int x)
{
	for (int i = a[x].size() - 1; ~i; --i)
	{
		int y = a[x][i];
		double sum = (sz[x] - 1 - sz[y]) / 2.0;
		ans[y] = ans[x] + sum + 1;
		dfs2(y);
	}
}
int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; ++i)a[i].clear();
		for (int i = 2; i <= n; ++i)
		{
			int x; scanf("%d", &x);
			a[x].push_back(i);
		}
		ans[1] = 1;
		dfs1(1);
		dfs2(1);
		for (int i = 1; i <= n; ++i)printf("%.12f\n", ans[i]);
	}
	return 0;
}
/*
【題意】
給你一棵樹，然後每個節點遍歷自己子節點的拓撲序是任意的（其訪問子節點的全排列等概率化）。
dfs帶有時間戳。問你，每個節點的訪問時間戳的期望是多少。

【型別】
樹的遍歷 概率均分思想

【分析】
這道題乍一看有些不知道如何入手。
難道我們要考慮所有全排列？
不過實際發現，所謂全排列規律均等，
其實也就意味著——
任意兩個兄弟節點的訪問概率都是均等的。
即：
每個節點有50%的概率在其兄弟節點的子樹之後訪問。

於是，我們求出每棵子樹的大小，
然後每個節點的訪問時間戳期望=父節點訪問時間戳期望+1+∑兄弟節點子樹大小/2
這道題就可以解決啦。

【時間複雜度&&優化】
O(n)

*/