考虑只有一个排列怎么做。有一个结论是答案为 \(n\ -\) 置换环个数，即每个环都会选择一个点不操作，其他点都操作。
接下来考虑两个排列，显然当 \(x\) 在 \(a\) 和 \(b\) 中都不操作，\(x\) 才能不操作。设 \(x\) 在 \(a\) 中所在环为 \(pa_x\)，在 \(b\) 中所在环为 \(pb_x\)，那么若 \(x\) 操作了，那么 \(pa_x\) 和 \(pb_x\) 的其他所有点都要操作。可以看成是二分图匹配，每条匹配边表示 \(x\) 不被操作，于是答案为 \(n\ -\) 最大匹配。
至于求方案，若 \(x\) 连的那条匹配边未满流，那么 \(x\) 就要被操作。
code
/*
p_b_p_b txdy
AThousandSuns txdy
Wu_Ren txdy
Appleblue17 txdy
*/
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 300100;
const int inf = 0x3f3f3f3f;
int n, a[maxn], b[maxn], head[maxn], len = 1, S, T, ntot;
struct edge {
int to, next, cap, flow;
} edges[maxn];
void add_edge(int u, int v, int c, int f) {
edges[++len].to = v;
edges[len].next = head[u];
edges[len].cap = c;
edges[len].flow = f;
head[u] = len;
}
struct dsu {
int fa[maxn];
void init() {
for (int i = 1; i <= n; ++i) {
fa[i] = i;
}
}
int find(int x) {
return fa[x] == x ? x : fa[x] = find(fa[x]);
}
void merge(int x, int y) {
x = find(x);
y = find(y);
if (x != y) {
fa[x] = y;
}
}
} d1, d2;
struct Dinic {
int d[maxn], cur[maxn];
bool vis[maxn];
void add(int u, int v, int c) {
add_edge(u, v, c, 0);
add_edge(v, u, 0, 0);
}
bool bfs() {
queue<int> q;
for (int i = 1; i <= ntot; ++i) {
d[i] = -1;
vis[i] = 0;
}
q.push(S);
d[S] = 0;
vis[S] = 1;
while (q.size()) {
int u = q.front();
q.pop();
for (int i = head[u]; i; i = edges[i].next) {
edge &e = edges[i];
if (!vis[e.to] && e.cap > e.flow) {
vis[e.to] = 1;
d[e.to] = d[u] + 1;
q.push(e.to);
}
}
}
return vis[T];
}
int dfs(int u, int a) {
if (!a || u == T) {
return a;
}
int flow = 0, f;
for (int &i = cur[u]; i; i = edges[i].next) {
edge &e = edges[i];
if (d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[i ^ 1].flow -= f;
flow += f;
a -= f;
if (!a) {
break;
}
}
}
return flow;
}
int solve() {
int flow = 0;
while (bfs()) {
for (int i = 1; i <= ntot; ++i) {
cur[i] = head[i];
}
flow += dfs(S, inf);
}
return flow;
}
} solver;
void solve() {
scanf("%d", &n);
d1.init();
d2.init();
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
d1.merge(i, a[i]);
}
for (int i = 1; i <= n; ++i) {
scanf("%d", &b[i]);
d2.merge(i, b[i]);
}
S = n * 2 + 1;
T = ntot = n * 2 + 2;
for (int i = 1; i <= n; ++i) {
solver.add(d1.find(i), d2.find(i) + n, 1);
}
for (int i = 1; i <= n; ++i) {
if (d1.fa[i] == i) {
solver.add(S, i, 1);
}
if (d2.fa[i] == i) {
solver.add(i + n, T, 1);
}
}
int flow = solver.solve();
printf("%d\n", n - flow);
for (int i = 1; i <= n; ++i) {
if (!edges[i * 2].flow) {
printf("%d ", i);
}
}
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}