P4168 [Violet] 蒲公英 Solution

Description

给定序列 $a=(a_1,a_2,\cdots ,a_n)$，有 $q$ 个查询 $(l,r)$.
对每个查询求 $a_{l\cdots r}$ 的众数，若有多个取最小的.
强制在线.
真实的 $l=(l^{\prime }+\text{lastans}-1)\mathrm{mod}n)+1,r=(r^{\prime }+\text{lastans}-1)\mathrm{mod}n)+1$，若 $l>r$ 则交换两者.

Limitations

$1\le n\le 4\times 10^4$
$1\le n\le 5\times 10^4$
$1\le a_i\le 10^9$
$1\le l\le r\le n$

Solution

区间众数无法用线段树维护，但 $n,m$ 不大，可以用分块.
首先 $a_i$ 很大，需要先离散化.
然后借助一个桶 $\text{cnt}$ 预处理 ${\text{pre}}_{i,j}$ 表示 $a_i$ 在第 $1\sim j$ 块的出现次数.
还有 ${\text{mode}}_{i,j}$ 表示第 $i\sim j$ 块的众数.

考虑查询，如果 $[l,r]$ 在单块内或相邻块间，我们直接暴力.
否则，我们统计两端散块内的 $a_i$ 在 中间整块内 以及 两端 的出现次数，存入 $\text{cnt}$ 中，并求出这部分的众数.
然后查询整块间的众数，以及 $[l,r]$ 间这个数的出现次数（整块散块分开算），然后判断并更新答案.
有几个坑：

• 算 ${\text{mode}}_{i,j}$ 前要先继承 ${\text{mode}}_{i,j-1}$ 的结果.
• 答案要转化为离散化前的.
• $\text{cnt}$ 用完要清空，可以只清空用过的部分.

取 $B=\sqrt{n}$，时间复杂度 $O((n+m)\sqrt{n})$.

Code

$4.09\text{KB},1.29\text{s},9.62\text{MB}\text{(in total, C++20 with O2)}$
fastio 删了

#include <bits/stdc++.h>
using namespace std;using i64 = long long;
using ui64 = unsigned long long;
using i128 = __int128;
using ui128 = unsigned __int128;
using f4 = float;
using f8 = double;
using f16 = long double;template<class T>
bool chmax(T &a, const T &b){if(a < b){ a = b; return true; }return false;
}template<class T>
bool chmin(T &a, const T &b){if(a > b){ a = b; return true; }return false;
}namespace fastio {}
using fastio::read;
using fastio::write;struct Block {int n, v, size, blocks;vector<int> a, belong, L, R, cnt;vector<vector<int>> pre, mode; inline Block() {}inline Block(const vector<int>& _a, int _v) : v(_v), a(_a) {n = a.size();size = (int)sqrt(n);blocks = (n + size - 1) / size;belong.resize(n);L.resize(blocks);R.resize(blocks);for (int i = 0; i < blocks; i++) {L[i] = i * size;R[i] = std::min(L[i] + size, n) - 1;for (int j = L[i]; j <= R[i]; j++) belong[j] = i;}pre.resize(blocks, vector<int>(v, 0));mode.resize(blocks, vector<int>(blocks, -1));cnt.resize(v);init();}// Prepareinline void init() {for (int i = 0; i < n; i++) for (int j = belong[i]; j < blocks; j++) pre[j][a[i]]++;for (int i = 0; i < blocks; i++) {clear();for (int j = i; j < blocks; j++) {if (j > i) mode[i][j] = mode[i][j - 1];for (int k = L[j]; k <= R[j]; k++) {cnt[a[k]]++;update(mode[i][j], a[k], cnt[a[k]]);}}}clear();}// Updateinline void update(int& x, int y, int c) {if (x == -1 || (c > cnt[x]) || (c == cnt[x] && y < x)) x = y;}// Clearinline void clear() { cnt.assign(v, 0); }inline void clear(int l, int r) {for (int i = l; i <= r; i++) cnt[a[i]] = 0;}// Countinline int count_block(int bl, int br, int k) { return pre[br - 1][k] - pre[bl][k];}inline int count_part(int l, int r, int k) {int res = 0;for (int i = l; i <= r; i++) res += (a[i] == k);return res;}// Add to the bucketinline void add(int l, int r, int bl, int br) {for (int i = l; i <= r; i++)if (!cnt[a[i]]) cnt[a[i]] += count_block(bl, br, a[i]);}// Update the answerinline void get_part(int& ans, int l, int r) {for (int i = l; i <= r; i++) {cnt[a[i]]++;update(ans, a[i], cnt[a[i]]);}}// Queryinline int query(int l, int r) {const int bl = belong[l], br = belong[r];if (br - bl <= 2) {int ans = -1;get_part(ans, l, r), clear(l, r);return ans;}int ans = -1;add(l, R[bl], bl, br), add(L[br], r, bl, br);get_part(ans, l, R[bl]), get_part(ans, L[br], r);int k = mode[bl + 1][br - 1];int cntk = (count_block(bl, br, k) + count_part(l, R[bl], k) + count_part(L[br], r, k));update(ans, k, cntk);clear(l, R[bl]), clear(L[br], r);return ans;}
};signed main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);const int n = read<int>(), m = read<int>();vector<int> a(n);for (int i = 0; i < n; i++) a[i] = read<int>();auto disc = a;sort(disc.begin(), disc.end());disc.erase(unique(disc.begin(), disc.end()), disc.end());for (int i = 0; i < n; i++) {a[i] = lower_bound(disc.begin(), disc.end(), a[i]) - disc.begin();}Block blk(a, disc.size());for (int i = 0, l, r, lst = 0; i < m; i++) {l = read<int>(), r = read<int>();l = (l + lst - 1) % n;r = (r + lst - 1) % n;if (l > r) swap(l, r);write(lst = disc[blk.query(l, r)]);putchar_unlocked('\n');}return 0;
}