20240929

Tree Journey

猜结论

#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 1e5 + 5;int n, k, ans, dep[N];vector<int> g[N];void dfs(int u, int f) {dep[u] = dep[f] + 1;for (auto v : g[u]) {if (v == f) {continue;}dfs(v, u);}if (dep[u] - 1 <= k) {ans = max(ans, min(n, dep[u] + (k - dep[u] + 1) / 2));}
}signed main() {cin >> n >> k;for (int i = 1, u, v; i < n; i++) {cin >> u >> v;g[u].push_back(v);g[v].push_back(u);}dfs(1, 0);cout << ans;return 0;
}

Infinity Card Decks

我们显然只用考虑 \(a[i] > b[i]\) 的数对,假设现在考虑第 \(j\) 张牌是否能打断无限牌组.
设\(suma, sumb\)表示\(a[i],b[i]\)之和.
\(M - (suma - sumb) - b[j] < 0\)
\(M - (suma - sumb) - a[j] < 0\)
我们显然会发现这个是单调的,所以就可以了

#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 1e6 + 5, INF = 1e18;int n, m, sum[N], suma[N], sumb[N], p[N], a[N], b[N], res[N], ans, pos[N], pos1[N];int tr[3][N * 4];void Merge(int i) {tr[1][i] = tr[1][i * 2] + tr[1][i * 2 + 1];tr[2][i] = max(tr[2][i * 2], tr[2][i * 2 + 1]);
}void build(int i, int l, int r) {if (l == r) {tr[1][i] = 0;if (a[l] > b[l]) {tr[2][i] = b[l];}else {tr[2][i] = a[l];}return ;}int mid = (l + r) >> 1;build(i * 2, l, mid);build(i * 2 + 1, mid + 1, r);Merge(i);
}int query_max(int i, int l, int r, int x, int y) {if (x > y) {return -INF;}if (l > y || r < x) {return -INF;}if (l >= x && r <= y) {return tr[2][i];}int mid = (l + r) >> 1;return max(query_max(i * 2, l, mid, x, y), query_max(i * 2 + 1, mid + 1, r, x, y));
}int query_sum(int i, int l, int r, int x, int y) {if (x > y) {return 0;}if (l > y || r < x) {return 0;}if (l >= x && r <= y) {return tr[1][i];}int mid = (l + r) >> 1;return query_sum(i * 2, l, mid, x, y) + query_sum(i * 2 + 1, mid + 1, r, x, y);
}void modify(int i, int l, int r, int p, int x) {if (l == r) {tr[1][i] += x;return ;}int mid = (l + r) >> 1;if (mid >= p) modify(i * 2, l, mid, p, x);else modify(i * 2 + 1, mid + 1, r, p, x);Merge(i);
}signed main() {ios::sync_with_stdio(0);cin.tie(0);cin >> n >> m;for (int i = 1; i <= n; i++) {cin >> a[i];}for (int i = 1; i <= n; i++) {cin >> b[i];}build(1, 1, n);for (int i = 1; i <= n; i++) {sum[i] = sum[i - 1] + (a[i] - b[i]);suma[i] = suma[i - 1];sumb[i] = sumb[i - 1];if (a[i] > b[i]) {suma[i] += a[i];sumb[i] += b[i];}p[i] = sum[i];}sort(p + 1, p + n + 1);for (int i = 0; i <= n; i++) {pos[i] = lower_bound(p + 1, p + n + 1, sum[i]) - p;pos1[i] = upper_bound(p + 1, p + n + 1, sum[i]) - p;}for (int i = 1, j = 1; i <= n; i++) {while (j <= n && m - ((suma[j] - suma[i - 1]) - (sumb[j] - sumb[i - 1])) - query_max(1, 1, n, i, j) >= 0) {j++;}res[i] = j;}for (int i = 1, j = 1; i <= n; i++) {while (j < res[i]) {modify(1, 1, n, pos[j], 1);j++;}ans += (n - res[i] + 1);ans += query_sum(1, 1, n, pos1[i - 1], n);if (i <= res[i]) {modify(1, 1, n, pos[i], -1);}}cout << (n * (n + 1) / 2) - ans;return 0;
}