[cf contest697] D - Puzzles

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转载最新推荐文章于 2024-02-07 07:00:00 发布 · 182 阅读

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原文链接：http://www.cnblogs.com/whc200305/p/8044701.html

本文详细解析了CFContest697 D题“Puzzles”，介绍了使用随机深度优先搜索（DFS）算法解决该问题的方法，并提供了一段C++代码实现。通过对子树中各节点的期望访问时间进行计算，展示了如何处理树形结构中的随机遍历问题。

[cf contest697] 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'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.
$\text{[math]}$
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 

哎，还是太菜。

不解释题意了。

像我数学那么菜。。。还是要学学奇技淫巧。

在一棵以x为根的子树中，对于x的子节点u和v，u有一半的概率rand到v前面访问，也有一半概率rand到后面。

所以u对v的贡献就是size[v]/2.

所以，E[y]=E[x]+1+(size[x]-size[y]-1)/2。

code：

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <vector>
 4 using namespace std;
 5 
 6 const int N=100005;
 7 int n,size[N]; double E[N];
 8 vector <int> e[N];
 9 
10 void dfs_size (int x,int p) {
11     size[x]=1;
12     for (int s=e[x].size(),i=0,y; i<s; ++i) {
13         if (e[x][i]==p) continue;
14         y=e[x][i];
15         dfs_size(y,p);
16         size[x]+=size[y];
17     }
18 }
19 void dfs_E (int x,int p) {
20     for (int s=e[x].size(),i=0,y; i<s; ++i) {
21         if (e[x][i]==p) continue;
22         y=e[x][i];
23         E[y]=E[x]+1+1.0*(size[x]-size[y]-1)/2;
24         dfs_E(y,x);
25     }
26 }
27 
28 int main () {
29     int x;
30     scanf("%d",&n);   
31     for (int i=2; i<=n; ++i) {
32         scanf("%d",&x);
33         e[x].push_back(i);
34     }
35     dfs_size(1,0);
36     E[1]=1;
37     dfs_E(1,0);
38     for (int i=1; i<=n; ++i) {
39         printf("%.6lf ",E[i]);
40     }
41     return 0;
42 }

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转载于:https://www.cnblogs.com/whc200305/p/8044701.html