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Progressive hints first, then the full explanation and implementation when you're ready to cash out.
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Hints
Open only as much as you need to keep the solve alive.
Normalize the labels by their positions in . Then is just , and a valid subtree rooted at must be some value interval that is also contiguous in the transformed .
For a start position in , scan forward while also scanning from the same root's position in . A common block ends exactly when the two scanned sets become equal.
Blocks that reverse order between the two permutations are stuck together. Model this as entanglement: a maximal entangled component behaves like one collapsed block in the outside problem.
Once every nontrivial entangled component is collapsed, the outside has identical DFS order in both sequences. Counting trees for one fixed preorder is equivalent to counting legal placements of DFS closing brackets.
The collapsed nontrivial components are forced leaves in that bracket DP. Use for forced leaves, for normal vertices, and transition with suffix sums over the number of currently open vertices.
First, normalize the labels. Relabel every vertex by its position in . Then Arietta's order becomes
and Tortinita's order becomes one permutation .
For any subtree rooted at , its vertices are contiguous in both DFS orders, and the root is first inside that subtree. Since is now sorted, such a subtree must be exactly a value interval . In , the same values must appear contiguously starting at .
So the basic object is a common interval with a fixed first vertex.
Entanglement
Suppose two child subtrees of the same parent appear in different orders in the two DFS sequences. Then those two children are entangled. They cannot be separated independently, because whichever tree you pick has to explain both DFS orders at once.
Take connected components under this entanglement relation. A maximal entangled component, together with everything inside its subtrees, forms one interval in both DFS orders. If its size is greater than , then in every valid tree the whole component must hang under one same parent. So we may collapse it into a single forced leaf in the outer problem, then solve the inside recursively.
That is the main trick. Crossing order is not extra freedom; it is a handcuff.
After all nontrivial entangled components are collapsed, the remaining outer DFS order is identical in both sequences.
Counting One Fixed Preorder
Now solve this smaller problem:
We have one preorder list of vertices. Some vertices are forced leaves. How many rooted forests realize this preorder?
Think of DFS as brackets:
The openings are fixed by the preorder. We choose where the closings go. If a vertex is forced to be a leaf, it must close immediately after it opens.
Let vec[i] be:
0 for a forced leaf,1 for a normal vertex.Let be the number of ways after processing a prefix, where vertices from that prefix are still open. When we add a vertex, it contributes vec[i] open vertices after its own immediate behavior. Then before the next vertex, we may close any suffix of currently open vertices.
So the transition is
That is just a suffix sum, so counting a block of length costs .
Building The Recursive Blocks
Process positions of from right to left. For a position , let be the position of the same label in .
Scan lengths and maintain the difference between the two sets
and
Whenever this difference is empty, we found a common block ending at that position in .
Keep active block starts that have not been attached to a larger block yet. The largest common block starting at absorbs all active starts inside it; they become children in the decomposition tree.
A boundary before child is clean if is already a common block. Clean boundaries continue the identical-order chain. The other pieces are collapsed entangled components, which become forced leaves in the bracket DP.
The implementation stores:
child[x]: recursive children of block x,cntForced[x]: forced collapsed leaves before the continuing child,go[x]: the one continuing child, or -1 if the chain ends here.Then DFS over this decomposition:
0 for each forced collapsed leaf,go[x] exists, append 1 and recurse into it without ending the current bracket DP,vec and multiply it into the answer,1 and recurse into that child.The set scans cost , and the bracket DPs also total . Memory is linear apart from the decomposition edges. This is exactly why the counting version uses : quadratic is fine; cubic gets cooked.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 998244353;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
int main() {
setIO();
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
vector<int> a(n), b(n), pb(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
--a[i];
}
for (int i = 0; i < n; i++) {
cin >> b[i];
--b[i];
pb[b[i]] = i;
}
vector<vector<int>> child(n + 1);
vector<int> cntForced(n + 1), go(n + 1, -1);
vector<char> active(n + 1, 0), good(n + 1, 0);
vector<int> diff(n, 0);
auto build_node = [&](int root, int ai, int bi) {
fill(good.begin(), good.end(), 0);
vector<int> touched;
int nonzero = 0;
int last = -1;
auto add = [&](int x, int delta) {
if (diff[x] == 0) {
touched.push_back(x);
nonzero++;
}
diff[x] += delta;
if (diff[x] == 0) nonzero--;
};
for (int len = 0; ai + len < n && bi + len < n; len++) {
add(a[ai + len], 1);
add(b[bi + len], -1);
if (nonzero == 0) {
good[ai + len] = 1;
last = ai + len;
}
}
for (int x : touched) diff[x] = 0;
cntForced[root] = 0;
go[root] = -1;
for (int j = ai + 1; j <= last; j++) {
if (!active[j]) continue;
child[root].push_back(j);
active[j] = 0;
if (good[j - 1]) {
cntForced[root]++;
go[root] = j;
} else {
go[root] = -1;
}
}
if (go[root] != -1) cntForced[root]--;
};
for (int i = n - 1; i >= 0; i--) {
build_node(i, i, pb[a[i]]);
active[i] = 1;
}
fill(good.begin(), good.end(), 0);
vector<int> touched;
int nonzero = 0;
auto add_root = [&](int x, int delta) {
if (diff[x] == 0) {
touched.push_back(x);
nonzero++;
}
diff[x] += delta;
if (diff[x] == 0) nonzero--;
};
for (int i = 0; i < n; i++) {
add_root(a[i], 1);
add_root(b[i], -1);
if (nonzero == 0) good[i] = 1;
}
for (int x : touched) diff[x] = 0;
int root = n;
cntForced[root] = 0;
go[root] = -1;
for (int i = 0; i < n; i++) {
if (!active[i]) continue;
child[root].push_back(i);
active[i] = 0;
if (i == 0 || good[i - 1]) {
cntForced[root]++;
go[root] = i;
} else {
go[root] = -1;
}
}
if (go[root] != -1) cntForced[root]--;
vector<int> vec;
vector<ll> dp(n + 2), ndp(n + 2);
auto count_same_order = [&]() -> ll {
int m = (int) vec.size();
fill(dp.begin(), dp.begin() + m + 2, 0);
dp[0] = 1;
for (int x : vec) {
fill(ndp.begin(), ndp.begin() + m + 2, 0);
for (int d = 0; d <= m; d++) {
if (dp[d]) ndp[d + x] = (ndp[d + x] + dp[d]) % MOD;
}
ll suf = 0;
for (int d = m; d >= 0; d--) {
suf += ndp[d];
if (suf >= MOD) suf -= MOD;
dp[d] = suf;
}
}
vec.clear();
return dp[0];
};
ll ans = 1;
auto dfs = [&](auto&& self, int v) -> void {
for (int i = 0; i < cntForced[v]; i++) vec.push_back(0);
if (go[v] == -1) {
ans = ans * count_same_order() % MOD;
} else {
vec.push_back(1);
self(self, go[v]);
}
for (int u : child[v]) {
if (u == go[v]) continue;
vec.push_back(1);
self(self, u);
}
};
dfs(dfs, root);
cout << ans << '\n';
}
}#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 998244353;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
int main() {
setIO();
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
vector<int> a(n), b(n), pb(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
--a[i];
}
for (int i = 0; i < n; i++) {
cin >> b[i];
--b[i];
pb[b[i]] = i;
}
vector<vector<int>> child(n + 1);
vector<int> cntForced(n + 1), go(n + 1, -1);
vector<char> active(n + 1, 0), good(n + 1, 0);
vector<int> diff(n, 0);
auto build_node = [&](int root, int ai, int bi) {
fill(good.begin(), good.end(), 0);
vector<int> touched;
int nonzero = 0;
int last = -1;
auto add = [&](int x, int delta) {
if (diff[x] == 0) {
touched.push_back(x);
nonzero++;
}
diff[x] += delta;
if (diff[x] == 0) nonzero--;
};
for (int len = 0; ai + len < n && bi + len < n; len++) {
add(a[ai + len], 1);
add(b[bi + len], -1);
if (nonzero == 0) {
good[ai + len] = 1;
last = ai + len;
}
}
for (int x : touched) diff[x] = 0;
cntForced[root] = 0;
go[root] = -1;
for (int j = ai + 1; j <= last; j++) {
if (!active[j]) continue;
child[root].push_back(j);
active[j] = 0;
if (good[j - 1]) {
cntForced[root]++;
go[root] = j;
} else {
go[root] = -1;
}
}
if (go[root] != -1) cntForced[root]--;
};
for (int i = n - 1; i >= 0; i--) {
build_node(i, i, pb[a[i]]);
active[i] = 1;
}
fill(good.begin(), good.end(), 0);
vector<int> touched;
int nonzero = 0;
auto add_root = [&](int x, int delta) {
if (diff[x] == 0) {
touched.push_back(x);
nonzero++;
}
diff[x] += delta;
if (diff[x] == 0) nonzero--;
};
for (int i = 0; i < n; i++) {
add_root(a[i], 1);
add_root(b[i], -1);
if (nonzero == 0) good[i] = 1;
}
for (int x : touched) diff[x] = 0;
int root = n;
cntForced[root] = 0;
go[root] = -1;
for (int i = 0; i < n; i++) {
if (!active[i]) continue;
child[root].push_back(i);
active[i] = 0;
if (i == 0 || good[i - 1]) {
cntForced[root]++;
go[root] = i;
} else {
go[root] = -1;
}
}
if (go[root] != -1) cntForced[root]--;
vector<int> vec;
vector<ll> dp(n + 2), ndp(n + 2);
auto count_same_order = [&]() -> ll {
int m = (int) vec.size();
fill(dp.begin(), dp.begin() + m + 2, 0);
dp[0] = 1;
for (int x : vec) {
fill(ndp.begin(), ndp.begin() + m + 2, 0);
for (int d = 0; d <= m; d++) {
if (dp[d]) ndp[d + x] = (ndp[d + x] + dp[d]) % MOD;
}
ll suf = 0;
for (int d = m; d >= 0; d--) {
suf += ndp[d];
if (suf >= MOD) suf -= MOD;
dp[d] = suf;
}
}
vec.clear();
return dp[0];
};
ll ans = 1;
auto dfs = [&](auto&& self, int v) -> void {
for (int i = 0; i < cntForced[v]; i++) vec.push_back(0);
if (go[v] == -1) {
ans = ans * count_same_order() % MOD;
} else {
vec.push_back(1);
self(self, go[v]);
}
for (int u : child[v]) {
if (u == go[v]) continue;
vec.push_back(1);
self(self, u);
}
};
dfs(dfs, root);
cout << ans << '\n';
}
}