Fix a Tree 利用并查集合并树，解决树中存在环的问题

Fix a Tree
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
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Description

A tree is an undirected connected graph without cycles.

Let's consider a rooted undirected tree with n vertices, numbered 1 through n.
There are many ways to represent such a tree. One way is to create an array with n integers p1, p2, ..., pn,
where pi denotes a parent of vertex i (here, for convenience a root is considered its
own parent).


 For this rooted tree the array p is [2, 3, 3, 2].
Given a sequence p1, p2, ..., pn, one is able to restore a tree:

There must be exactly one index r that pr = r. A vertex r is
a root of the tree. 
For all other n - 1 vertices i, there is an edge between vertex i and
vertex pi. 
A sequence p1, p2, ..., pn is called valid if the described procedure
generates some (any) rooted tree. For example, for n = 3 sequences (1,2,2), (2,3,1) and (2,1,3) are
not valid.

You are given a sequence a1, a2, ..., an, not necessarily valid. Your
task is to change the minimum number of elements, in order to get a valid sequence. Print the minimum number of changes and an example of a valid sequence after that number of changes. If there are many valid sequences achievable in the minimum number of changes,
print any of them.

Input

The first line of the input contains an integer n (2 ≤ n ≤ 200 000) — the number of vertices in the tree.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n).

Output

In the first line print the minimum number of elements to change, in order to get a valid sequence.

In the second line, print any valid sequence possible to get from (a1, a2, ..., an) in
the minimum number of changes. If there are many such sequences, any of them will be accepted.

Sample Input

Input

4
2 3 3 4

Output

1
2 3 4 4

Input

5
3 2 2 5 3

Output

0
3 2 2 5 3

Input

8
2 3 5 4 1 6 6 7

Output

2
2 3 7 8 1 6 6 7

Hint

In the first sample, it's enough to change one element. In the provided output, a sequence represents a tree rooted in a vertex 4 (because p4 = 4),
which you can see on the left drawing below. One of other correct solutions would be a sequence 2 3 3 2, representing a tree rooted in vertex 3 (right drawing below).
On both drawings, roots are painted red.



In the second sample, the given sequence is already valid.

题目大意，给你一些子树（先默认存在环的图为树），把它合并成一棵树，这个题目中需要解决的是，首先要判断该树种是否有环，如果有环的话，需要先把环处理掉，把带有环的树拆开一条边并把它合并到其他的树上

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define inf 0x3f3f3f3f
#define maxn 200002
#define maxm 10001
using namespace std;

int n;
int f[maxn],t[maxn];//f是合并后的树，t是存储一开始时的树
int book[maxn];//book为1代表改点需要改变
int getf(int i)
{
if(i==f[i])
return i;
else
return f[i]=getf(f[i]);
}

int merge(int x, int y)
{
int u=getf(x);
int v=getf(y);
if(u!=v)
{
f[v]=u;
return 1;
}
return 0;
}

int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&t[i]);
f[i]=i;//初始化树
}

int pre=-1,cnt=0;//pre是上一次合并后的节点
//先处理掉图中存在的多个环的问题
for(int i=1;i<=n;i++)
{
//存在环，比如1-2-3-1
if(!merge(t[i],i) && t[i]!=i)//该判断条件用来判断该树中是否存在环
{
cnt++;
book[i]=1;//该节点需要改变指向，所以赋值为1
if(pre!=-1)
{
merge(i,pre);
}
pre=i;//pre改变为上次改变的节点的位置
}
}

for(int i=1;i<=n;i++)
{
if(i==f[i])//最后合并一下各种子树
{
book[i]=1;
cnt++;
if(pre!=-1)
{
merge(i,pre);
}
pre=i;
}
}
printf("%d\n",cnt-1);
for(int i=1;i<=n;i++)
{
if(book[i])
printf("%d ",f[i]);
else printf("%d ",t[i]);
}
return 0;
}