CF363Ddiv2 D Fix a Tree

晚上寫殘了，來補一下。

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:

1. There must be exactly one index r that pr = r. A vertex r is a root of the tree.
2. 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 (becausep4 = 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.

題意：給定一個圖（可能有環），問最少改變幾個結點能讓森林變成一棵樹。

思路：因爲要變成一棵樹，那肯定只有一個根，所以加入這個圖裏有n個樹或環，那就要把其中的n-1個樹或環接到另外一個樹下面。

所以這個題就是求這個圖有幾個根，然後把所有的根移到同一個根下面就好了。

先用並查集找根，如果已經有現成的樹根了，就把別的環和根移到這下面，如果沒有現成的樹根，也就是說，都是環，那麼就把一個環變成樹根，別的在連過來。

代碼：

//************************************************************************//
//*Author : Handsome How                                                 *//
//************************************************************************//
//#pragma comment(linker, "/STA    CK:1024000000,1024000000")
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#include <ctime>
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define foreach(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define fur(i,a,b) for(int i=(a);i<=(b);i++)
#define furr(i,a,b) for(int i=(a);i>=(b);i--)
#define cl(a) memset((a),0,sizeof(a))
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#ifdef HandsomeHow
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define dbg(x) cout << #x << " = " << x << endl
#else
#define debug(...)
#define dbg(x)
#endif
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair <int, int> pii;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const int mod=1000000007;
const double pi=acos(-1);
inline void gn(long long&x){
    int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
int gcd(int a,int b){return a? gcd(b%a,a):b;}
ll powmod(ll a,ll x,ll mod){ll t=1ll;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}
// (づ°ω°)づe★
//-----------------------------------------------------------------
const int maxn = 222222;
int p[maxn], v[maxn];
int n,rroot,subroot;
int findf(int x){
	if(x == p[x]) return x;
	int t = findf(p[x]);
	p[x] = t;
	return t;
}

void merge(int a, int b){
	a = findf(a); b = findf(b);
	if(a != b) p[a] = b;
}

int main(){
#ifdef HandsomeHow
    //freopen("E:\\data.in","r",stdin);
    //freopen("E:\\data.out","w",stdout);
    time_t beginttt = clock();
#endif
	gn(n);
	rroot = -1;
	fur(i,1,n) p[i] = i;
	fur(i,1,n){
		gn(v[i]);
		if(v[i] == i) rroot = i;
		merge(i,v[i]);
	}
	fur(i,1,n) if(findf(i) == i) subroot = i;
	if(rroot == -1) rroot = subroot;	 //是否有現成的樹根 
	int ans = 0;
	if(v[rroot] != rroot){
		ans++;
		v[rroot] = rroot;
	}
	for(int i = 1; i <= n; ++i){
		if(findf(i) == i && i != rroot){
			ans++;
			v[i] = rroot;
		}
	}
	printf("%d\n",ans);
	fur(i,1,n) printf("%d ",v[i]);
	puts("");
#ifdef HandsomeHow
	time_t endttt = clock();
    debug("time: %d\n",(int)(endttt - beginttt));
#endif
	return 0;
}