用C++写There are n folders in a computer, indexed from 1 to n . The folder indexed 1 is the root folder, and each folder may contain several subfolders, forming a tree structure. You can perform several "cut" operations, each cutting a source folder into a target folder. Note that the target folder cannot be the same as the source folder, nor can it be the source folder's subfolder, subfolder's subfolder, and so on. Obviously, the root folder cannot be cut, and it will always be the root folder. Find out that at least how many "cut" operations should be performed so that each folder contains at most one subfolder. Input The first line contains a single integer n (1≤n≤105) — the number of folders. The second line of each test case contains n−1 space separated integers p2,p3,⋯,pn (1≤pi≤n) — pi represents the index of the parent folder of folder i . Output Output the minimum number of "cut" operations performed. Note that there may not be a need to perform "cut" operation. Please output 0 in this case.
时间: 2024-01-16 19:02:10 浏览: 40
以下是使用 C++ 解决该问题的代码:
```cpp
#include <iostream>
#include <vector>
using namespace std;
const int MAXN = 1e5 + 5;
vector<int> G[MAXN];
int ans;
void dfs(int u, int fa) {
int cnt = 0;
for (int v : G[u]) {
if (v == fa) continue;
dfs(v, u);
cnt++;
}
ans += max(cnt - 1, 0);
}
int main() {
int n;
cin >> n;
for (int i = 2; i <= n; i++) {
int x;
cin >> x;
G[x].push_back(i);
}
dfs(1, 0);
cout << ans << endl;
return 0;
}
```
该问题可以用一个简单的 DFS 解决。我们首先读入每个节点的父亲节点,然后建立一棵树。对于每个节点,我们统计它的子节点个数,记为 $cnt$。如果 $cnt \geq 2$,则需要将其中的 $cnt - 1$ 个子节点切断,使得每个节点最多只有一个子节点。因此,我们在 DFS 的过程中累加所有需要切断的次数,最终输出即可。
相关问题
请问这段文字中讲的带标注的文件怎么用来计算查准率(3)Annotated_UNICODE Files in the folder STNO_UNICODE are annotated and put here. Each file is annotated by three annotators. Therefore, there are three folders for files of each topic.
标注文件可以用来计算查准率(precision)指标。在这个任务中,每个文件都是由三名注释者进行标注的。可以将这三个标注版本合并,生成一个标注结果文件,然后将其与系统提取出的结果文件进行比较,计算出查准率。具体来说,查准率是指系统提取的结果中正确的比例,即被系统判定为正确的结果占整个系统提取结果的比例。可以通过比较系统提取结果中的每个答案是否与标注结果相同来计算查准率。
f.path for f in os.scandir(folder_path) if f.is_dir()如何修改
如果你想修改这行代码,可以根据你的需求进行调整,以下是一些可能的修改方案:
1. 如果你想获取指定文件夹下的所有文件(不包括文件夹),可以将 `if f.is_dir()` 改为 `if f.is_file()`,这样就只会返回文件路径了。
2. 如果你想在获取文件夹路径时排除某些文件夹,可以加入判断条件,例如:
```python
excluded_folders = ['__pycache__', 'tmp']
folder_paths = [f.path for f in os.scandir(folder_path) if f.is_dir() and f.name not in excluded_folders]
```
这样就会排除名字为 `__pycache__` 和 `tmp` 的文件夹了。
3. 如果你想获取指定文件夹下的所有文件和文件夹(包括隐藏文件和文件夹),可以将 `os.scandir()` 改为 `os.listdir()`,例如:
```python
all_paths = [os.path.join(folder_path, f) for f in os.listdir(folder_path)]
```
这样就会返回指定文件夹下的所有文件和文件夹路径了。