LG P4119 [Ynoi2018] 未来日记 Solution

Description

给定序列 $a=(a_1,a_2,\cdots ,a_n)$，有 $m$ 个操作分两种：

• $\mathrm{replace}(l,r,x,y)$：将 $a_l\sim a_r$ 中的所有 $x$ 替换为 $y$.
• $\mathrm{kth}(l,r,k)$：求 $a_l\sim a_r$ 中的第 $k$ 小值.

Limitations

$1\le n,m,a_i,x,y\le 10^5$
$1\le l\le r\le n$
$1\text{s},512\text{MB}$

Solution

很奇怪的操作，考虑给序列 $a$ 分块.
由于要查询第 $k$ 小，再给值域 $[1,10^5]$ 分块.
预处理两个数组：

• $f_{i,j}$：第 $1\sim i$ 个序列块中，$j$ 的出现次数.
• $g_{i,j}$：第 $1\sim i$ 个序列块中，第 $j$ 个值域块内数的的出现次数.

并在每块挂上一个 vector 存储该块的操作，重构块时，遍历 vector 计算出 ${\text{to}}_i$ 表示 $i$ 这个数值最终会变成什么，然后直接在 $a$ 上修改，并把 vector 清空.

考虑修改，对于整块，我们直接把操作塞入对应块的 vector 中；对于散块，我们暴力替换.
当然 $f$ 和 $g$ 也要更新，所以替换时需要记录 $t_i$ 表示第 $i$ 块内的替换次数，在所有块修改完后更新这两个数组.

接着考虑查询，我们先对每个 $i$ 求出第 $i$ 个值域块内数在 $a_l\sim a_r$ 的出现次数，然后定出第 $k$ 小值所在的值域块.
接着再求出数值 $i$ 在 $a_l\sim a_r$ 的出现次数，再在上一步确定的值域块内找出真正的第 $k$ 小值.
此时 $f$ 和 $g$ 就发挥了作用，因为要求出连续整块内一个数（或一个值域块内数）的出现次数.（散块就直接暴力装桶）

至此就想完了，然而本题很毒瘤（又难写又卡常），注意以下几点：

• 块长取 $350$ 左右，太小会爆空间.
• 注意区分序列块和值域块.
• $x=y$ 的修改直接忽略.
• 不要开太多 vector，要加快读快写.
• 注意优化清空过程.

Code

$5.05\text{KB},6.25\text{s},204.35\text{MB}\text{(in total, C++20 with O2)}$
常数很大，最大点要 $0.84\text{s}$.

#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 {}      // By pystraf
using fastio::read;
using fastio::write;constexpr int B = 350, V = 1e5 + 10, M = (V + B - 1) / B;
using pii = pair<int, int>;namespace pool {constexpr int L = 4e7 + 10;int P[L], *ptr = P;inline int* alloc(int n) {int* res = ptr;ptr += n;return res;}
}
using pool::alloc;signed main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);const int n = read<int>(), q = read<int>();const int blocks = (n + B - 1) / B;int* a = alloc(n);for (int i = 0; i < n; i++) a[i] = read<int>();int *bel = alloc(V), *to = alloc(V), *L = alloc(blocks), *R = alloc(blocks), *tmp = alloc(V);vector<int*> f(blocks), g(blocks);vector<vector<pii>> mdf(blocks);for (int i = 0; i < blocks; i++) f[i] = alloc(V), g[i] = alloc(M);auto init = [&]() {for (int i = 0; i < V; i++) {bel[i] = i / B;to[i] = i;}for (int i = 0; i < blocks; i++) {L[i] = i * B;R[i] = min(L[i] + B, n) - 1;if (i > 0) {for (int j = 0; j < V; j++) f[i][j] = f[i - 1][j];for (int j = 0; j < M; j++) g[i][j] = g[i - 1][j];}for (int j = L[i]; j <= R[i]; j++) {f[i][a[j]]++;g[i][bel[a[j]]]++;}}};auto refact = [&](int b) {vector<int> vec;vector<pii> & opr = mdf[b];reverse(opr.begin(), opr.end());for (auto & [x, y] : opr) {to[x] = to[y];vec.push_back(x);vec.push_back(y);}for (int i = L[b]; i <= R[b]; i++) a[i] = to[a[i]];for (auto x : vec) to[x] = x;opr.clear();};auto brute_replace = [&](int b, int l, int r, int x, int y) {refact(b);for (int i = l; i <= r; i++)if (a[i] == x) {a[i] = y;tmp[b]++;}};auto block_replace = [&](int b, int x, int y) {mdf[b].emplace_back(x, y);tmp[b] = f[b][x] - f[b - 1][x];};auto update = [&](int b, int x, int y) {for (int i = b; i < blocks; i++) {if (i > 0) tmp[i] += tmp[i - 1];f[i][x] -= tmp[i], f[i][y] += tmp[i];g[i][bel[x]] -= tmp[i], g[i][bel[y]] += tmp[i];}};auto replace = [&](int l, int r, int x, int y) {if (x == y) return;const int bl = bel[l], br = bel[r];for (int i = 0; i < blocks; i++) tmp[i] = 0;if (bl == br) brute_replace(bl, l, r, x, y);else {brute_replace(bl, l, R[bl], x, y);brute_replace(br, L[br], r, x, y);for (int i = bl + 1; i < br; i++) block_replace(i, x, y);}update(bl, x, y);};auto count = [&](int bl, int br, int fl, int fr, int k, int sum, const vector<int>& vec) -> pii {if (fl == -1) {for (int i = 0; i < M; i++) tmp[i] = (bl ^ br) ? (g[br - 1][i] - g[bl][i]) : 0;for (auto & x : vec) tmp[bel[x]]++;for (int i = 0; i < M; i++) {sum += tmp[i];if (sum >= k) return pii(i, sum - tmp[i]);}}else {for (int i = fl; i <= fr; i++) tmp[i] = (bl ^ br) ? (f[br - 1][i] - f[bl][i]) : 0;for (auto & x : vec) tmp[x]++;for (int i = fl; i <= fr; i++) {sum += tmp[i];if (sum >= k) return pii(i, 0);}}return pii(-1, 0);};auto add = [&](int l, int r, vector<int>& vec) {for (int i = l; i <= r; i++) vec.push_back(a[i]);};auto kth = [&](int l, int r, int k) -> int {vector<int> vec;const int bl = bel[l], br = bel[r];if (bl == br) {refact(bl);add(l, r, vec);auto [blk, sum] = count(bl, bl, -1, -1, k, 0, vec);if (blk == -1) return -1;const int fl = blk * B, fr = min(fl + B, V) - 1;auto [res, _] = count(bl, bl, fl, fr, k, sum, vec);return res;}refact(bl), refact(br);add(l, R[bl], vec), add(L[br], r, vec);auto [blk, sum] = count(bl, br, -1, -1, k, 0, vec);if (blk == -1) return -1;const int fl = blk * B, fr = min(fl + B, V) - 1;auto [res, _] = count(bl, br, fl, fr, k, sum, vec);return res;};init();for (int i = 0, op, l, r, x, y; i < q; i++) {op = read<int>(), l = read<int>(), r = read<int>(), x = read<int>();l--, r--;if (op == 1) y = read<int>(), replace(l, r, x, y);else {write(kth(l, r, x));putchar_unlocked('\n');}}return 0;
}