本题链接

从叶子（尚未访问过的节点）往上找其最远的祖先，若能找到根，则合题；若找到环或找到与树不连通的祖先，则将祖先的父亲指向根。

#include <bits/stdc++.h>
namespace FastIO {
    char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh = '\n';
    int p, p3 = -1;

    void read() {}

    void print() {}

    inline int getc() {
        return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
    }

    inline void flush() {
        fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;
    }

    template<typename T, typename... T2>
    inline void read(T &x, T2 &... oth) {
        int f = 0;
        x = 0;
        char ch = getc();
        while (!isdigit(ch)) {
            if (ch == '-')
                f = 1;
            ch = getc();
        }
        while (isdigit(ch)) {
            x = x * 10 + ch - 48;
            ch = getc();
        }
        x = f ? -x : x;
        read(oth...);
    }

    template<typename T, typename... T2>
    inline void print(T x, T2... oth) {
        if (p3 > 1 << 20)
            flush();
        if (x < 0)
            buf2[++p3] = 45, x = -x;
        do {
            a[++p] = x % 10 + 48;
        } while (x /= 10);
        do {
            buf2[++p3] = a[p];
        } while (--p);
        buf2[++p3] = hh;
        print(oth...);
    }
} // namespace FastIO
#define read FastIO::read
#define print FastIO::print
//======================================
using namespace std;
const int maxn=2e5+10;
typedef long long ll;
int n,m,dep[maxn],root,cnt,fa[maxn],ans,in[maxn],t;
void work(int u) {
    dep[u] = ++cnt;
    while (!dep[fa[u]]) {
        u = fa[u];
        dep[u]=cnt;
    }
    if (dep[fa[u]] == cnt) {
        if (root == 0) root = u;
        if (fa[u]!=root){
            ans++;
            fa[u] = root;
        }
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
    //freopen("2.txt", "w", stdout);
#endif
    //======================================
    read(n);
    for (int i = 1; i <= n; i++) {
        read(fa[i]);
        if (fa[i] == i) root = i;
    }
    for (int i = 1; i <= n; i++) {
        if (!dep[i]) {
            work(i);
        }
    }
    print(ans);
    FastIO::hh = ' ';
    for (int i = 1; i <= n; i++) {
        print(fa[i]);
    }
    //======================================
    FastIO::flush();
    return 0;
}