CF280C-Game on Tree

题目:

题目描述:

Momiji has got a rooted tree, consisting of $n$ nodes. The tree nodes are numbered by integers from $1$ to $n$ . The root has number $1$ . Momiji decided to play a game on this tree.

The game consists of several steps. On each step, Momiji chooses one of the remaining tree nodes (let’s denote it by $v$ ) and removes all the subtree nodes with the root in node $v$ from the tree. Node $v$ gets deleted as well. The game finishes when the tree has no nodes left. In other words, the game finishes after the step that chooses the node number $1$ .

Each time Momiji chooses a new node uniformly among all the remaining nodes. Your task is to find the expectation of the number of steps in the described game.

输入格式:

The first line contains integer $n$ $(1<=n<=10^{5})$ — the number of nodes in the tree. The next $n-1$ lines contain the tree edges. The $i$ -th line contains integers $a_{i}$ , $b_{i}$ $(1<=a_{i},b_{i}<=n; a_{i}≠b_{i})$ — the numbers of the nodes that are connected by the $i$ -th edge.

It is guaranteed that the given graph is a tree.

输出格式:

Print a single real number — the expectation of the number of steps in the described game.

The answer will be considered correct if the absolute or relative error doesn’t exceed $10^{-6}$ .

样例:

样例输入1:

1
2
3

2
1 2

样例输出1:

1
2

1.50000000000000000000

样例输入2:

1
2
3
4

3
1 2
1 3

样例输出2:

1
2

2.00000000000000000000

思路:

实现:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52

// Problem: CF280C Game on Tree
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF280C
// Memory Limit: 250 MB
// Time Limit: 1000 ms
// Author: Ybw051114
//
// Powered by CP Editor (https://cpeditor.org)

#include "ybwhead/ios.h"
const int maxn = 1e5 + 10;
struct edge
{
    int v, nxt;
} e[maxn << 1];
int head[maxn], tot;
void __ADD(int u, int v)
{
    e[++tot] = {v, head[u]};
    head[u] = tot;
}
void add(int u, int v)
{
    __ADD(u, v);
    __ADD(v, u);
}
double ans;
int n;
void dfs(int u, int fa, int dep)
{
    ans += (double)1.0 / dep;
    for (int i = head[u]; i; i = e[i].nxt)
    {
        int v = e[i].v;
        if (v == fa)
            continue;
        dfs(v, u, dep + 1);
    }
}
int main()
{
    yin >> n;
    for (int i = 1; i < n; i++)
    {
        int u, v;
        yin >> u >> v;
        add(u, v);
    }
    dfs(1, 0, 1);
    yout << ans << endl;
    return 0;
}