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The node is special as hell: once it ever becomes a leaf, it can never be removed, because no larger leaf exists. So root the tree at and think of the process as deleting leaves upward toward the root.
A leaf stays a leaf until it is removed. Therefore, if is the current maximum leaf, is “protected” until some larger vertex becomes a leaf. The distinct values inserted into are strictly increasing.
Suppose the current protected maximum is , and the next new protected maximum is . To make a leaf, every vertex below in the tree rooted at must be deleted while is still protecting the process.
Define as the maximum label in the rooted subtree of , excluding itself. Then a transition is possible exactly when
If some descendant of is bigger than , it would become the next maximum first. If is inside that subtree, you literally cannot delete it. Same inequality catches both issues.
Now it is just path-counting in a DAG over labels. Let be the maximum initial leaf. Set , and for :
Use a Fenwick tree for these range sums. Finally, you may jump from last record to root iff every vertex outside the root-child branch containing has label .
First, notice the endgame: vertex can never be removed once it becomes a leaf, because it is the largest possible leaf. So the final remaining vertex is always , and is always in .
Root the tree at . Every deletion now removes some currently available rooted-subtree frontier vertex.
Another key fact: once a vertex becomes a leaf, it stays a leaf until removed. If it is the maximum-index leaf, it cannot be removed. Therefore the distinct values added to form a strictly increasing sequence:
The first value is forced: it is the maximum-index leaf in the original tree. Call it .
For a vertex , define
or if is a leaf.
Suppose the current record leaf is , and the next new record is .
For to become a leaf, every vertex in the rooted subtree of , except itself, must be deleted before appears. During this whole phase, is still the maximum protected leaf.
So every deleted vertex below must have label . If some descendant had label , that descendant would become the next maximum before . If the descendant had label , then it is the protected vertex itself and cannot be deleted. Either way, the transition dies.
Thus we need:
Also records strictly increase, so . Therefore:
This is both necessary and sufficient. If , just prune the whole subtree below while protects everything. Then becomes a leaf, and since , it becomes the next record. No magic, no hand-wavy bullshit.
The final jump is special because is the root. To make a leaf, we do not need to delete the branch containing the current protected record . We only need all other root-child branches gone.
Let be the child of that lies on the path from to . For each root child , define:
Then we can finish from to iff the maximum label outside 's branch is smaller than :
After that, becomes a leaf, joins , and protects the rest of the process until the tree is fully destroyed.
Since every valid set corresponds to exactly one increasing record sequence, we just count paths in this implicit DAG.
Let be the number of valid record sequences ending at record , before reaching .
The start is forced:
For every from to :
That is a range sum over labels, so use a Fenwick tree.
Finally:
If , then is already the maximum initial leaf, so the only possible set is .
Rooting the tree and computing all values takes . The DP uses Fenwick operations, so each test case is:
with memory. Across all test cases, this easily fits since .
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
const int MOD = 998244353;
struct Fenwick {
int n;
vector<int> bit;
Fenwick(int n = 0) { init(n); }
void init(int n_) {
n = n_;
bit.assign(n + 1, 0);
}
void add(int idx, int val) {
for (; idx <= n; idx += idx & -idx) {
bit[idx] += val;
if (bit[idx] >= MOD) bit[idx] -= MOD;
}
}
int sumPrefix(int idx) const {
int res = 0;
for (; idx > 0; idx -= idx & -idx) {
res += bit[idx];
if (res >= MOD) res -= MOD;
}
return res;
}
int rangeSum(int l, int r) const {
if (l > r) return 0;
int res = sumPrefix(r) - sumPrefix(l - 1);
if (res < 0) res += MOD;
return res;
}
};
int main() {
setIO();
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
vector<vector<int>> g(n + 1);
vector<int> deg(n + 1, 0);
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
deg[u]++;
deg[v]++;
}
int start = 1;
for (int i = 1; i <= n; i++) {
if (deg[i] == 1) start = max(start, i);
}
if (start == n) {
cout << 1 << '\n';
continue;
}
vector<int> parent(n + 1, 0), top(n + 1, 0), order;
order.reserve(n);
parent[n] = -1;
order.push_back(n);
for (int i = 0; i < (int)order.size(); i++) {
int v = order[i];
for (int to : g[v]) {
if (to == parent[v]) continue;
parent[to] = v;
top[to] = (v == n ? to : top[v]);
order.push_back(to);
}
}
// mx[v] = maximum label in the rooted subtree of v, excluding v.
vector<int> mx(n + 1, 0);
for (int i = n - 1; i >= 0; i--) {
int v = order[i];
if (v == n) continue;
int p = parent[v];
mx[p] = max(mx[p], v);
mx[p] = max(mx[p], mx[v]);
}
vector<int> dp(n + 1, 0);
Fenwick fw(n);
dp[start] = 1;
fw.add(start, 1);
for (int v = start + 1; v < n; v++) {
dp[v] = fw.rangeSum(mx[v] + 1, v - 1);
if (dp[v]) fw.add(v, dp[v]);
}
vector<int> branchMax(n + 1, 0);
int best1 = 0, best2 = 0, bestChild = 0;
for (int c : g[n]) {
branchMax[c] = max(c, mx[c]);
if (branchMax[c] > best1) {
best2 = best1;
best1 = branchMax[c];
bestChild = c;
} else if (branchMax[c] > best2) {
best2 = branchMax[c];
}
}
int ans = 0;
for (int v = start; v < n; v++) {
if (!dp[v]) continue;
int c = top[v];
int outsideMax = (c == bestChild ? best2 : best1);
if (outsideMax < v) {
ans += dp[v];
if (ans >= MOD) ans -= MOD;
}
}
cout << ans << '\n';
}
}#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
const int MOD = 998244353;
struct Fenwick {
int n;
vector<int> bit;
Fenwick(int n = 0) { init(n); }
void init(int n_) {
n = n_;
bit.assign(n + 1, 0);
}
void add(int idx, int val) {
for (; idx <= n; idx += idx & -idx) {
bit[idx] += val;
if (bit[idx] >= MOD) bit[idx] -= MOD;
}
}
int sumPrefix(int idx) const {
int res = 0;
for (; idx > 0; idx -= idx & -idx) {
res += bit[idx];
if (res >= MOD) res -= MOD;
}
return res;
}
int rangeSum(int l, int r) const {
if (l > r) return 0;
int res = sumPrefix(r) - sumPrefix(l - 1);
if (res < 0) res += MOD;
return res;
}
};
int main() {
setIO();
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
vector<vector<int>> g(n + 1);
vector<int> deg(n + 1, 0);
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
deg[u]++;
deg[v]++;
}
int start = 1;
for (int i = 1; i <= n; i++) {
if (deg[i] == 1) start = max(start, i);
}
if (start == n) {
cout << 1 << '\n';
continue;
}
vector<int> parent(n + 1, 0), top(n + 1, 0), order;
order.reserve(n);
parent[n] = -1;
order.push_back(n);
for (int i = 0; i < (int)order.size(); i++) {
int v = order[i];
for (int to : g[v]) {
if (to == parent[v]) continue;
parent[to] = v;
top[to] = (v == n ? to : top[v]);
order.push_back(to);
}
}
// mx[v] = maximum label in the rooted subtree of v, excluding v.
vector<int> mx(n + 1, 0);
for (int i = n - 1; i >= 0; i--) {
int v = order[i];
if (v == n) continue;
int p = parent[v];
mx[p] = max(mx[p], v);
mx[p] = max(mx[p], mx[v]);
}
vector<int> dp(n + 1, 0);
Fenwick fw(n);
dp[start] = 1;
fw.add(start, 1);
for (int v = start + 1; v < n; v++) {
dp[v] = fw.rangeSum(mx[v] + 1, v - 1);
if (dp[v]) fw.add(v, dp[v]);
}
vector<int> branchMax(n + 1, 0);
int best1 = 0, best2 = 0, bestChild = 0;
for (int c : g[n]) {
branchMax[c] = max(c, mx[c]);
if (branchMax[c] > best1) {
best2 = best1;
best1 = branchMax[c];
bestChild = c;
} else if (branchMax[c] > best2) {
best2 = branchMax[c];
}
}
int ans = 0;
for (int v = start; v < n; v++) {
if (!dp[v]) continue;
int c = top[v];
int outsideMax = (c == bestChild ? best2 : best1);
if (outsideMax < v) {
ans += dp[v];
if (ans >= MOD) ans -= MOD;
}
}
cout << ans << '\n';
}
}