人员分组以后如何计算人数的问题，怎么采用 C 程序的代码编写的过程的方式加以实现的

Question

Problem Description
One day dandelion and her friends decides to play outside. As there are so many people , they decide to divide into several parts . Dandelion wants to know how many friends will be together with her. The way they use to divide is : everyone can choose one people to join in her group, Such as if A chooses B to join in her group , then B will leave the group she is in now , and joins in the group that A is in . One can also choose some people who has already in her group. All of dandelion抯 friends are expressed as an integer 2,3... n and dandelion is expressed as 1.The choosing work begins from dandelion then the friend who are signed as 2 ?until the friend who are signed as n .

Input
Each case begins with an integer n, (0<n<=1000000) stands for the number of people.
Then follows n lines , the i+1 line contains an integer p, stands for people who are signed as i chooses the one who are signed as p to join in her group.

Output
The output contains one line for each data set : the number of friends will be together with dandelion at last .

Sample Input
3
3
3
1
3
2
1
2

Sample Output
2
0

Answer 1

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这是一个典型的图论问题，我们可以使用邻接矩阵来表示这个图。在C++中，我们可以创建一个二维数组，其中每个元素代表图中的一个节点，并且值为1或0，表示是否被选择。

以下是一个简单的实现：

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    
    int graph[1005][1005];
    // 初始化邻接矩阵
    for(int i = 0; i <= n; i++) {
        for(int j = 0; j <= n; j++) {
            graph[i][j] = 0;
        }
    }

    while(cin >> n) {
        int a, b;
        cin >> a >> b;
        
        graph[a-1][b-1] = 1;
        graph[b-1][a-1] = 1;
    }

    int res = 0;
    for(int i = 0; i <= n; i++) {
        for(int j = 0; j <= n; j++) {
            if(graph[i][j]) {
                res++;
            }
        }
    }
    cout << res << endl;
    return 0;
}

在这个程序中，我们首先读取n个数据点，然后遍历每一个数据点，将它们加入到邻接矩阵中。最后，我们遍历所有的节点，并检查它们是否已经加入了某个节点，如果是，则我们将res加一。

这只是一个基本的实现，你可以根据你的需求进行修改和优化。例如，你可能想要添加一些错误处理，或者使用更高效的算法来找出答案。