2024/09/30 模拟赛总结
\(0+0+42+40\)，T1在写正解的时候突然比赛还有1分钟结束，然后把 freopen 注释的暴力在最后几秒交了上去
唐氏 xor-hashing，首先博弈游戏很简单，如果有一个数的出现次数是奇数则先手必胜，否则先手必败
那么先手必败的条件就是路径上所有边权都是两两配对的，即异或和为 \(0\)。那么钦定点 \(1\) 为根，求出它到每个点的异或和，用 map 存下每个和的出现次数。答案为 \({n \choose 2} - \sum {\text{mp}_i \choose 2}\)
因为边权太小，异或可能会挂掉，所以需要随机一个较大值来异或防止错误
// BLuemoon_
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int kMaxN = 5e5 + 5;
LL cnt, s[kMaxN];
vector<pair<int, LL>> g[kMaxN];
int t, n;
map<LL, LL> h, ans;
mt19937_64 rd(time(0) + 3247 + 1508);
void DFS(int u, int fa) {
ans[s[u]]++;
for (auto [v, w] : g[u]) {
if (v != fa) {
s[v] = s[u] ^ h[w], DFS(v, u);
}
}
}
int main() {
freopen("game.in", "r", stdin), freopen("game.out", "w", stdout);
for (cin >> t; t; t--, ans.clear()) {
cin >> n;
for (int i = 1, u, v, w; i < n; i++) {
cin >> u >> v >> w, g[u].push_back({v, w}), g[v].push_back({u, w}), (!h[w]) && (h[w] = rd());
}
s[1] = 0, DFS(1, 0), cnt = 1ll * n * (n - 1) / 2;
for (auto [u, v] : ans) {
cnt -= v * (v - 1) / 2;
}
cout << cnt << '\n';
for (int i = 1; i <= n; i++) {
g[i].clear();
}
}
return 0;
}
显然跳跃形式一定是形如 \(l_1 \rightarrow r_1 \rightarrow l_1 \dots l_1 \rightarrow r_2 \rightarrow l_2 \rightarrow r_2 \rightarrow l_2 \rightarrow r_3 \dots r_m \rightarrow l_m\)，其中 \(l_i \le r_1 \le l_2 \le r_2 \le \dots \le l_m \le r_m\)
令 \(dp_i\) 为跳到 \(i\)，最小跳跃次数以及分数。然后枚举上一个段来直接大力转移即可
// BLuemoon_
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int kMaxN = 1e3 + 5;
LL t, n, k, a[kMaxN << 1], s[kMaxN << 1], mx[kMaxN << 1], ans, e[kMaxN << 1], vis[kMaxN << 1], g[kMaxN << 1];
queue<LL> q;
LL Calc(LL x, LL y) { return s[max((x - 1) % n + 1, (y - 1) % n + 1)] - s[min((x - 1) % n + 1, (y - 1) % n + 1) - 1]; }
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
freopen("jump.in", "r", stdin), freopen("jump.out", "w", stdout);
for (cin >> t; t; t--, ans = 0) {
cin >> n >> k;
for (int i = 1; i <= n; i++) {
cin >> a[i], s[i] = s[i - 1] + a[i];
}
fill(mx + 1, mx + 2 * n + 1, -1e18), fill(e + 1, e + 2 * n + 1, 1e18), fill(vis + 1, vis + 2 * n + 1, 0), fill(g + 1, g + 2 * n + 1, -1e18);
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
mx[i] = max(mx[i], (s[j] - s[i - 1]) << 1), mx[j + n] = max(mx[j + n], (s[j] - s[i - 1]) << 1);
}
}
for (e[1] = g[1] = 0, q.push(1); !q.empty();) {
LL u = q.front();
q.pop(), vis[u]++;
if (vis[u] == (n << 1 | 1)) {
continue;
}
for (int i = 1; i <= n << 1; i++) {
if ((u <= n && i >= u + n) || (i <= n && u >= i + n)) {
LL v = e[u], w = g[u], tmp = Calc(i, u);
if (w + tmp < 0 && mx[u] <= 0) {
continue;
}
if (w + tmp < 0) {
LL p = (mx[u] - 1 - w - tmp) / mx[u];
v += p << 1, w += p * mx[u];
}
w += tmp, v++;
if (v < e[i] || v == e[i] && w > g[i]) {
e[i] = v, g[i] = w, q.push(i);
}
}
}
}
for (int i = 1; i <= n << 1; i++) {
if (e[i] <= k) {
ans = max({ans, g[i], (k - e[i]) / 2 * mx[i]});
if (e[i] != k) {
for (int j = 1; j <= n << 1; j++) {
LL tmp = Calc(i, j);
(tmp >= 0) && (ans = max(ans, g[i] + (k - 1 - e[i]) / 2 * mx[i] + tmp));
}
}
}
}
cout << ans << '\n';
}
return 0;
}
纯人类智慧，进行一遍 DFS 所当前子树节点数量大于 \(B\)，则直接割开。最后剩下遗留的子树合起来一定小于 \(3B\)，直接分为一组即可
// BLuemoon_
#include <bits/stdc++.h>
using namespace std;
const int kMaxN = 1e4 + 5;
int n, b, tot, ans[kMaxN], sz[kMaxN], m[kMaxN];
vector<int> g[kMaxN], s[kMaxN];
void DFS(int u, int fa) {
for (int v : g[u]) {
if (v != fa) {
DFS(v, u), sz[u] += sz[v], (s[u].size() < s[v].size()) && (swap(s[u], s[v]), 0);
for (int tmp : s[v]) {
s[u].push_back(tmp);
}
if (sz[u] >= b) {
m[++tot] = u;
for (int tmp : s[u]) {
ans[tmp] = tot;
}
s[u].clear(), sz[u] = 0;
}
}
}
sz[u]++, s[u].push_back(u);
if (sz[u] >= b) {
m[++tot] = u;
for (int tmp : s[u]) {
ans[tmp] = tot;
}
s[u].clear(), sz[u] = 0;
}
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
freopen("mainland.in", "r", stdin), freopen("mainland.out", "w", stdout);
cin >> n >> b;
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);
for (int tmp : s[1]) {
ans[tmp] = tot;
}
cout << tot << '\n';
for (int i = 1; i <= n; i++) {
cout << (ans[i] ? ans[i] : tot) << " \n"[i == n];
}
for (int i = 1; i <= tot; i++) {
cout << m[i] << " \n"[i == tot];
}
return 0;
}
要写 FHQ，理解思路但懒得写，先咕咕咕写完了
显然区间右移就是交换区间，可以用 FHQ 先割再连解决
令一个节点的 tag 为该子树内是否存在二元上升子序列，用子树最大值最小值就可以维护
那么我们类似地定义 \(mx^{\prime},mn^{\prime}\)，前者定义为子树内最大的值满足存在以这个值结尾的二元上升子序列，后者定义类似
首先有 \(mx^{\prime}\ge\max\{mx_{l}^{\prime},mx_{r}^{\prime}\}\)，因为可能两个节点一个在左一个在右
右边的点一定是 \(mx_{r}^{\prime}\)，所以我们只需要找到左子树最大的小于 \(mx_{r}^{\prime}\) 的值，为了优化，我们只需要维护不满足条件的值即可
// BLuemoon_
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const LL kMaxN = 1.2e5 + 5;
LL n, q, cnt, root, l, r, k, rt1, rt2, rt3, rt4, rt5, rt6, val[kMaxN], sz[kMaxN], ls[kMaxN], rs[kMaxN], mn[kMaxN], mx[kMaxN], L[kMaxN], R[kMaxN];
bool ans[kMaxN];
mt19937 Rand(time(0) / 2 + 3247 + 1508);
LL New(LL x) {
sz[++cnt] = 1, val[cnt] = mn[cnt] = mx[cnt] = x;
L[cnt] = 1e18, R[cnt] = -1e18;
return ans[cnt] = 0, cnt;
}
LL Pre(LL x, LL y) {
if (x == 0) {
return -1e18;
}
return val[x] >= y ? Pre(rs[x], y) : max(Pre(ls[x], y), val[x]);
}
LL Suf(LL x, LL y) {
if (x == 0) {
return 1e18;
}
return val[x] <= y ? Suf(ls[x], y) : min(Suf(rs[x], y), val[x]);
}
void Update(LL x) {
sz[x] = sz[ls[x]] + sz[rs[x]] + 1, ans[x] = ans[ls[x]] | ans[rs[x]] | (L[ls[x]] < max(val[x], mx[rs[x]]) || min(val[x], mn[ls[x]]) < R[rs[x]] || mn[ls[x]] < val[x] && val[x] < mx[rs[x]]);
mn[x] = min({val[x], mn[ls[x]], mn[rs[x]]}), mx[x] = max({val[x], mx[ls[x]], mx[rs[x]]}), L[x] = min(L[ls[x]], L[rs[x]]), R[x] = max(R[ls[x]], R[rs[x]]);
(val[x] < mx[rs[x]]) && (R[x] = max(R[x], val[x])), R[x] = max(R[x], Pre(ls[x], max(mx[rs[x]], val[x])));
(val[x] > mn[ls[x]]) && (L[x] = min(L[x], val[x])), L[x] = min(L[x], Suf(rs[x], min(mn[ls[x]], val[x])));
}
void Split(LL p, LL v, LL &x, LL &y) {
if (p == 0) {
return x = y = 0, void();
}
if (v == 0) {
return x = 0, y = p, void();
}
(v <= sz[ls[p]] ? y : x) = p, Split(v <= sz[ls[p]] ? ls[p] : rs[p], v - (sz[ls[p]] + 1) * (v > sz[ls[p]]), v <= sz[ls[p]] ? x : rs[p], v <= sz[ls[p]] ? ls[p] : y);
return Update(p), void();
}
int Merge(int x, int y) {
if (x == 0 || y == 0) {
return x + y;
}
if (Rand() % (sz[x] + sz[y]) < sz[x]) {
return rs[x] = Merge(rs[x], y), Update(x), x;
}
return ls[y] = Merge(x, ls[y]), Update(y), y;
}
void pr(bool pr) {
cout << (pr ? "YES" : "NO") << '\n';
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
freopen("permutation.in", "r", stdin), freopen("permutation.out", "w", stdout);
cin >> n, mx[root] = R[root] = -1e18, mn[root] = L[root] = 1e18;
for (int i = 1, x; i <= n; i++) {
cin >> x, root = Merge(root, New(x));
}
for (cin >> q; q; q--) {
cin >> l >> r >> k, rt1 = rt2 = rt3 = rt4 = rt5 = rt6 = 0;
Split(root, r, rt1, rt2), Split(rt1, r - k, rt3, rt4), Split(rt3, l - 1, rt5, rt6), root = Merge(Merge(rt5, rt4), Merge(rt6, rt2)), pr(ans[root]);
}
return 0;
}
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