发现每一个点都只有一个出边，所以形成了一个基环树森林。

记录每一个点下一条点和上一条点，大分讨即可。

注意利用以前的信息，否则时间复杂度会退化到 $O(n^2)$。

注意特判点不在环上的情况。

时间复杂度 $O(n)$(应该是)

// This guy is too lazy so that he/she write nothing.
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int nxt[N], deg[N], dis[N], step[N];
vector<int> pre[N];
bool vis[N], vis2[N], vis3[N];
signed main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int x;
        cin >> x;
        nxt[i] = x;
        pre[x].push_back(i);
        deg[i]++, deg[x]++;
    }
    memset(dis, 0x3f, sizeof dis);
    for (int i = 1; i <= n; i++)
        if (!vis[i])
        {
            int j = i, cnt = 0, laj;
            vector<int> tlist;
            while (!vis[j])
            {
                vis[j] = true;
                vis2[j] = true;
                tlist.push_back(j);
                step[j] = cnt++;
                laj = j;
                j = nxt[j];
            }
            if (vis2[j]) // 在环上
            {
                for (auto &j : tlist)
                    vis2[j] = false;
                int k = cnt - step[j];
                queue<int> q;
                tlist.clear();
                int tk = j;
                do
                {
                    q.push(j);
                    tlist.push_back(j);
                    dis[j] = k;
                    j = nxt[j];
                    vis2[j] = true;
                } while (tk != j);
                while (q.size())
                {
                    int f = q.front();
                    q.pop();
                    vis2[f] = false;
                    for (auto &j : pre[f])
                        if (dis[j] > dis[f] + 1)
                        {
                            dis[j] = dis[f] + 1;
                            if (!vis2[j])
                            {
                                vis2[j] = true;
                                q.push(j);
                                tlist.push_back(j);
                            }
                        }
                }
            }
            else // 不在环上
            {
                queue<int> q;
                q.push(j);
                for (auto &j : tlist)
                    vis2[j] = false;
                vis2[j] = true;
                tlist.clear();
                tlist.push_back(j);
                while (q.size())
                {
                    int f = q.front();
                    q.pop();
                    vis2[f] = false;
                    for (auto &j : pre[f])
                        if (dis[j] > dis[f] + 1)
                        {
                            dis[j] = dis[f] + 1;
                            if (!vis2[j])
                            {
                                vis2[j] = true;
                                q.push(j);
                                tlist.push_back(j);
                            }
                        }
                }
            }
            for (auto &j : tlist)
                vis2[j] = false;
        }
    for (int i = 1; i <= n; i++)
        cout << dis[i] << '\n';
    return 0;
}