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| 1 | +// Rozwiązanie zadania 'Strajki' z II etapu XXIV OI. |
| 2 | +// Autor: Paweł Putra |
| 3 | +// Złożoność czasowa: O(n + m * log(n)). |
| 4 | +// Złożoność pamięciowa: O(n). |
| 5 | + |
| 6 | +#include <iostream> |
| 7 | +#include <vector> |
| 8 | +#include <queue> |
| 9 | +#include <cassert> |
| 10 | +using namespace std; |
| 11 | +constexpr int MAXN = 1'000'005, INF = 1'000'000'005; |
| 12 | +int n, m; |
| 13 | +int vis[MAXN], strajk[MAXN], t[MAXN], L[MAXN], R[MAXN], par[MAXN]; |
| 14 | +vector<int> g[MAXN]; |
| 15 | + |
| 16 | +struct Fenwick { |
| 17 | + int n; |
| 18 | + vector<int> fw; |
| 19 | + |
| 20 | + Fenwick(int n) : n(n), fw(n, 0) {} |
| 21 | + |
| 22 | + void add(int x, int delta) { |
| 23 | + for (;x < n; x |= x + 1) |
| 24 | + fw[x] += delta; |
| 25 | + } |
| 26 | + int sum(int r) { |
| 27 | + int res = 0; |
| 28 | + for (; r >= 0; r = (r & (r + 1)) - 1) |
| 29 | + res += fw[r]; |
| 30 | + return res; |
| 31 | + } |
| 32 | + int sum(int l, int r) { |
| 33 | + if (l > r) return 0; |
| 34 | + return sum(r) - sum(l-1); |
| 35 | + } |
| 36 | +}; |
| 37 | + |
| 38 | +int32_t main() { |
| 39 | + ios_base::sync_with_stdio(0); |
| 40 | + cin >> n; |
| 41 | + |
| 42 | + for (int i = 1; i < n; i++) { |
| 43 | + int a, b; |
| 44 | + cin >> a >> b; |
| 45 | + assert(a < b); |
| 46 | + g[a].push_back(b); |
| 47 | + g[b].push_back(a); |
| 48 | + } |
| 49 | + |
| 50 | + int T = 0; |
| 51 | + |
| 52 | + // Ukorzeniamy drzewo w jedynce i numerujemy je w kolejności przechodzenia bfsem, |
| 53 | + // dzięki temu synowie wierzchołka zajmują spójny przedział numerów. |
| 54 | + queue<int> q; |
| 55 | + q.push(1); |
| 56 | + t[1] = ++T; |
| 57 | + while (!q.empty()) { |
| 58 | + int v = q.front(); |
| 59 | + q.pop(); |
| 60 | + L[v] = INF, R[v] = -INF; |
| 61 | + for (auto u : g[v]) { |
| 62 | + if (!t[u]) { |
| 63 | + t[u] = ++T; |
| 64 | + par[u] = v; |
| 65 | + L[v] = min(L[v], t[u]); |
| 66 | + R[v] = max(R[v], t[u]); |
| 67 | + q.push(u); |
| 68 | + } |
| 69 | + } |
| 70 | + } |
| 71 | + |
| 72 | + |
| 73 | + int spojne = 1; |
| 74 | + Fenwick fw(n+2); |
| 75 | + cin >> m; |
| 76 | + while (m--) { |
| 77 | + int v; |
| 78 | + cin >> v; |
| 79 | + if (v > 0) { |
| 80 | + assert(!strajk[v]); |
| 81 | + } else { |
| 82 | + v = -v; |
| 83 | + assert(strajk[v]); |
| 84 | + } |
| 85 | + |
| 86 | + strajk[v] = 1 - strajk[v]; |
| 87 | + int ile = (R[v] >= L[v] ? (R[v] - L[v] + 1) - fw.sum(L[v], R[v]) : 0); |
| 88 | + if (par[v] && !strajk[par[v]]) ile++; |
| 89 | + |
| 90 | + if (strajk[v]) { |
| 91 | + spojne += ile - 1; |
| 92 | + fw.add(t[v], 1); |
| 93 | + } else { |
| 94 | + spojne -= ile - 1; |
| 95 | + fw.add(t[v], -1); |
| 96 | + } |
| 97 | + |
| 98 | + cout << spojne << "\n"; |
| 99 | + } |
| 100 | +} |
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