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sqrt_decomposition.go
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/
sqrt_decomposition.go
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package copypasta
import (
"math"
"slices"
)
/* 根号算法 根号分治 Sqrt Decomposition
一种技巧：组合两种算法从而降低复杂度 O(n^2) -> O(n√n)
常用于图论或者某些数组统计类题目
参考 Competitive Programmer’s Handbook Ch.27
王悦同《根号算法——不只是分块》
todo 暴力美学——浅谈根号分治 https://www.luogu.com.cn/blog/Amateur-threshold/pu-li-mei-xue-qian-tan-gen-hao-fen-zhi
题目花样很多，下面举个例子
有 n 个对象，每个对象有一个「关于其他对象的统计量」ci（一个数、一个集合的元素个数，等等）
为方便起见，假设 ∑ci 的数量级和 n 一样，下面用 n 表示 ∑ci
当 ci > √n 时，这样的对象不超过 √n 个，暴力枚举这些对象之间的关系（或者，该对象与其他所有对象的关系），时间复杂度为 O(n) 或 O(n√n)。此乃算法一
当 ci ≤ √n 时，这样的对象有 O(n) 个，由于统计量 ci 很小，暴力枚举当前对象的统计量，时间复杂度为 O(n√n)。此乃算法二
这样，以 √n 为界，我们将所有对象划分成了两组，并用两个不同的算法处理
这两种算法是看待同一个问题的两种不同方式，通过恰当地组合（平衡）这两个算法，复杂度由 O(n^2) 降至 O(n√n)
例子是 https://codeforces.com/problemset/problem/1806/E
注意：**枚举时要做到不重不漏**
- [1714. 数组中特殊等间距元素的和](https://leetcode.cn/problems/sum-of-special-evenly-spaced-elements-in-array/)（会员题）
可以从这题上手 https://www.luogu.com.cn/problem/P3396
同 https://codeforces.com/contest/103/problem/D 2100
https://www.luogu.com.cn/problem/T279521?contestId=65460
- https://www.luogu.com.cn/blog/cyffff/solution-JRKSJ-Eltaw
LCP16 https://leetcode.cn/problems/you-le-yuan-de-you-lan-ji-hua/
https://codeforces.com/problemset/problem/1921/F 1900
https://codeforces.com/problemset/problem/797/E 2000
https://codeforces.com/problemset/problem/1207/F 2100
https://codeforces.com/problemset/problem/1806/E 2200
https://codeforces.com/problemset/problem/1968/G2 2200 也可以直接记忆化
https://codeforces.com/problemset/problem/425/D 2300
https://codeforces.com/problemset/problem/677/D 2300
https://codeforces.com/problemset/problem/1468/M 2300 或四元环
https://codeforces.com/problemset/problem/342/E 2400
https://codeforces.com/problemset/problem/506/D 2400
见下面的 floorDivide https://codeforces.com/problemset/problem/786/C 2400
大步+小步，有点分段打表的味道 https://codeforces.com/problemset/problem/1619/H 2400
https://codeforces.com/problemset/problem/1270/F 2600
https://codeforces.com/problemset/problem/1039/D 2800
https://codeforces.com/problemset/problem/1039/E 3400
四次方根 https://atcoder.jp/contests/abc293/tasks/abc293_f
https://leetcode.com/discuss/interview-question/3517350/
- Given a list of pairs {L,R} & an array.
Find out the total number of pairs (i, j) where
(arr[i], arr[j]) should be equal to exactly one of the given Q pairs. And i < j.
自动 O(n√n)
https://codeforces.com/problemset/problem/1790/F 2100
随机+分块
https://codeforces.com/contest/1840/problem/G2 2500
*/
// TIPS: n 的整数分拆中，不同数字的个数至多有 O(√n) 种
/* 分散层叠算法 Fractional Cascading
https://en.wikipedia.org/wiki/Fractional_cascading
https://www.zhihu.com/question/33776070/answer/59405602
https://www.luogu.com.cn/blog/DPair2005/fen-san-ceng-die-suan-fa-xue-xi-bi-ji
https://www.luogu.com.cn/problem/P6466
*/
/*
分块数据结构
https://oi-wiki.org/ds/decompose/
https://oi-wiki.org/ds/block-array/
Unrolled linked list https://en.wikipedia.org/wiki/Unrolled_linked_list
【推荐】https://www.luogu.com.cn/blog/220037/Sqrt1
浅谈基础根号算法——分块 https://www.luogu.com.cn/blog/deco/qian-tan-ji-chu-gen-hao-suan-fa-fen-kuai
todo https://www.csie.ntu.edu.tw/~sprout/algo2018/ppt_pdf/root_methods.pdf
区间 rank 的 N 种解法 https://www.luogu.com.cn/blog/Peterprpr/HunterNoHorse
浅谈归约矩乘 https://www.luogu.com.cn/blog/Ynoi/qian-tan-gui-yue-ju-cheng
为什么有些题只能用根号科技？https://www.luogu.com.cn/blog/include-c/sqrt-tech
题目推荐 https://cp-algorithms.com/data_structures/sqrt_decomposition.html#toc-tgt-8
好题 https://codeforces.com/problemset/problem/91/E
todo 动态逆序对 https://www.luogu.com.cn/problem/P3157 https://www.luogu.com.cn/problem/UVA11990
https://cp-algorithms.com/sequences/rmq.html
todo https://www.luogu.com.cn/problem/P3396
https://codeforces.com/problemset/problem/1207/F
https://codeforces.com/contest/455/problem/D
https://codeforces.com/problemset/problem/13/E 2700
https://www.luogu.com.cn/problem/P3203 [HN10] 弹飞绵羊
*/
func _(a []int) {
// 下标从 0 开始
type block struct {
l, r int // [l,r]
origin, sorted []int
//lazyAdd int
}
n := len(a)
blockSize := int(math.Sqrt(float64(n))) // 建议设为常量，避免超时 【提示】大一点可能更快一些
//blockSize := int(math.Sqrt(float64(n) * math.Log2(float64(n+1))))
blocks := make([]block, (n-1)/blockSize+1)
for i, v := range a {
j := i / blockSize
if i%blockSize == 0 {
blocks[j] = block{l: i, origin: make([]int, 0, blockSize)}
}
blocks[j].origin = append(blocks[j].origin, v)
}
for i := range blocks {
b := &blocks[i]
b.r = b.l + len(b.origin) - 1
b.sorted = append([]int(nil), b.origin...)
slices.Sort(b.sorted)
}
sqrtOp := func(l, r int, v int) { // [l,r], starts at 0
for i := range blocks {
b := &blocks[i]
if b.r < l {
continue
}
if b.l > r {
break
}
if l <= b.l && b.r <= r {
// do op on full block
} else {
// do op on part block
bl := max(b.l, l)
br := min(b.r, r)
for j := bl - b.l; j <= br-b.l; j++ {
// do b.origin[j]...
}
}
}
}
_ = sqrtOp
}
// 如果 f(i) 的计算结果近似 n/i
// 可以对 [1,n] 值域分治，如果区间内的结果都相同，则不再分治
// 时间复杂度类似整除分块 O(f(n)√n)
// https://codeforces.com/problemset/problem/786/C
func floorDivide(n int, f func(int) int) []int {
ans := make([]int, n+1)
var solve func(int, int)
solve = func(l, r int) {
if l > r {
return
}
resL, resR := f(l), f(r)
if resL == resR {
for i := l; i <= r; i++ {
ans[i] = resL
}
return
}
ans[l] = resL
ans[r] = resR
mid := (l + r) / 2
solve(l+1, mid)
solve(mid+1, r-1)
}
solve(1, n)
return ans[1:]
}