CodeForces 699D—— Fix a Tree(并查集判断环)

D. Fix a Tree

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

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

Note

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.

思路：首先，怎么用并查集判断时候有环？你在合并两个节点的时候，如果他们之前已经处在同一个连通分量里，那么把这两个节点合并的时候必定会产生环。

怎么建树？如果存在父节点指向自己的点，那么就把它作为这棵树的根。如果不存在，那么随便开一个环，把开环的节点作为根节点。

并查集的过程可以同时完成建树和查环的操作，如果合并这两个节点会形成环，那就开环后把一个节点指向根节点。如果不会，那就把他们合并。

#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <string>
#include <map>
#include <set>

using namespace std;

#define MAXN 200010
#define LEN 1000000
#define INF 1e9+7
#define MODE 1000000
typedef long long ll;

int par[MAXN];
void init(int n)
{
for(int i=1;i<=n;i++){
par[i]=i;
}
}

int find(int x){
if(par[x]==x)
return x;
else
return par[x]=find(par[x]);
}

void unite(int x,int y)
{
x=find(x);
y=find(y);
if(x==y)
return;
else
par[x]=y;
}

bool same(int x,int y)
{
return find(x)==find(y);
}

struct edge{
int u,v;
};
vector <edge> e;

int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
init(n);
int ans;
int root=-1;
for(int i=1;i<=n;i++)
{
edge temp;
temp.u=i;
scanf("%d",&temp.v);
e.push_back(temp);
if(temp.u==temp.v)
root=i;
}
int res=0;
for(int i=0;i<e.size();i++)
{
if(e[i].u!=root)
{
if(same(e[i].u,e[i].v))
{
res++;
if(root==-1)
root=e[i].u;
e[i].v=root;
}
else
unite(e[i].u,e[i].v);
}
}
printf("%d\n",res);
for(int i=0;i<e.size();i++)
{
printf("%d",e[i].v);
if(i!=e.size()-1)
printf(" ");
}
printf("\n");
}
}