PolandBall lives in a forest with his family. There are some trees in the forest. Trees are undirected acyclic graphs with k vertices and k - 1 edges, where k is some integer. Note that one vertex is a valid tree.
There is exactly one relative living in each vertex of each tree, they have unique ids from 1 to n. For each Ball i we know the id of its most distant relative living on the same tree. If there are several such vertices, we only know the value of the one with smallest id among those.
How many trees are there in the forest?
Input
The first line contains single integer n (1 ≤ n ≤ 104) — the number of Balls living in the forest.
The second line contains a sequence p1, p2, ..., pn of length n, where (1 ≤ pi ≤ n) holds and pi denotes the most distant from Ball i relative living on the same tree. If there are several most distant relatives living on the same tree, pi is the id of one with the smallest id.
It's guaranteed that the sequence p corresponds to some valid forest.
Hacking: To hack someone, you should provide a correct forest as a test. The sequence p will be calculated according to the forest and given to the solution you try to hack as input. Use the following format:
In the first line, output the integer n (1 ≤ n ≤ 104) — the number of Balls and the integer m (0 ≤ m < n) — the total number of edges in the forest. Then m lines should follow. The i-th of them should contain two integers ai and bi and represent an edge between vertices in which relatives ai and bi live. For example, the first sample is written as follows:

5 3
1 2
3 4
4 5
Output
You should output the number of trees in the forest where PolandBall lives.
Example
Input
5
2 1 5 3 3
Output
2
Input
1
1
Output
1
Note
In the first sample testcase, possible forest is: 1-2 3-4-5.
There are 2 trees overall.
In the second sample testcase, the only possible graph is one vertex and no edges. Therefore, there is only one tree.
这道题的题意，我也是看了半天才明白。
题意：给你n个数字，每一个数字表示与第i(1<=i<=n)个节点与那个节点相连，最后让你判断这个森林里有几棵树。
拿第一个样例来说，有五个节点，与第一个节点相连的为2,和第二个节点相连的为1,同理可得其他。
所以1和2在一棵树上，3，4，5在一棵树上。
这道题我用的是并查集。
主要的解题思路呢就是将每个节点和它相连的那个节点并起来，最后检查有多少个根（检查根呢，就是看那个节点的父亲是不是等于它本身）
代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define cls(x) memset(x,0,sizeof(x))
using namespace std;
const int MAXN=100010;
int par[MAXN];
void init(int n)
{
    for(int i=1;i<=n;i++)
     par[i]=i;
}
int find(int x)
{
    if(x==par[x])return x;
    return par[x]=find(par[x]);
}
void unite(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y)return;
    par[y]=x; 
}
int main()
{
    int n;
    scanf("%d",&n);
    init(n);
    for(int i=1;i<=n;i++)
     {
        int a;
        scanf("%d",&a);
        unite(i,a);
     }
    int ans=0;
    for(int i=1;i<=n;i++)
     if(i==par[i])ans++;//检查根节点个数
    printf("%d\n",ans); 
    return 0;
}