CodeForces 699D 并查集+判断环

Fix a Tree

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

Submit Status Practice CodeForces 699D

Appoint description: System Crawler  (Nov 20, 2016 9:42:44 PM)

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, ..., p_n, where p_i 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, ..., p_n, one is able to restore a tree:

1. There must be exactly one index r that p_r = 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 p_i.

A sequence p₁, p₂, ..., p_n 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, ..., a_n, 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, ..., a_n (1 ≤ a_i ≤ 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, ..., a_n) in the minimum number of changes. If there are many such sequences, any of them will be accepted.

Sample Input

Input

42 3 3 4

Output

12 3 4 4 

Input

53 2 2 5 3

Output

03 2 2 5 3 

Input

82 3 5 4 1 6 6 7

Output

22 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.

题意：

给你n个点，给你每个点的父节点的编号（类似并查集的描述）

让你求修改最少的点的父节点序号，让他变成一棵树，special judge。

解题思路：

给出的点有两种情况需要自习考虑：

1，一个点自环

2，有限条边形成环

那么，主要处理第二种，第一种的话可以直接合并到树里面

第二种：

采用加边的做法，把题目给的数组看成给你n条边，让你往图里面加，对于i如果边（pa[i],i）两端点在不同的连通子图里，那么就加入这条边。

否则就修改pa[i]=i;让他成为这个子图的root。

然后得到的就是一堆子树了。

然后合并这些子树就可以了，注意合并的时候尽量以没修改过的子树为根，保证修改次数最少。

代码：

#include<bits/stdc++.h>using namespace std;const int maxn = 2e5+5;int pa[maxn] ;int fa[maxn] ;bool vis[maxn] ;int findset(int x){return x==fa[x]?x:fa[x]=findset(fa[x]);}int main(){    int n;    while(~scanf("%d",&n)){        for(int i=1;i<=n;i++){            scanf("%d",&pa[i]);            fa[i]=i ;        }        memset(vis,false,sizeof(vis)) ;        for(int i=1;i<=n;i++){            //printf("i=%d\n",i);            int u=findset(i);            int v=findset(pa[i]);            //printf("u=%d v=%d\n",u,v);            if(u!=v){                //cnt++ ;                fa[u]=v ;                //pa[u]=v;            }else{                if(pa[i]!=i)vis[i]=true ;                pa[i]=i;            }        }        int root=0;        for(int i=1;i<=n;i++){            if(pa[i]==i){root=i;break ;}        }        for(int i=1;i<=n;i++){            if(pa[i]==i&&!vis[i]){root=i;break ;}        }        for(int i=1;i<=n;i++){            if(pa[i]==i&&i!=root){                pa[i]=root ;                vis[i]=true;            }        }        int cnt=0;        for(int i=1;i<=n;i++){            if(vis[i]){                //printf("i=%d\n",i);                cnt++ ;            }        }        printf("%d\n",cnt);        for(int i=1;i<=n;i++){            printf("%d",pa[i]);            if(i!=n)printf(" ");        }printf("\n");    }    return 0;}