P2801 教主的魔法 Solution

原创已于 2025-03-15 13:25:21 修改 · 837 阅读

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#算法 #c++ #数据结构 #分块

于 2025-02-23 20:36:18 首次发布

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Description

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

$add⁡(l,r,w)\operatorname{add}(l,r,w)$ ：对每个 $\in [l,r]$ 执行 $ai←ai+wa_i \gets a_i+w$ .
$query⁡(l,r,c)\operatorname{query}(l,r,c)$ ：求 $∑i=lr[ai≥c]\sum\limits_{i=l}^r [a_i \ge c]$ .

Limitations

$\le n \le 10^6$
$\le q \le 3000$
$\le w \le 1000$
$\le a_i,c \le 10^9$
$\text{s},128 \text{MB}$

Solution

线段树不好维护，考虑分块.
由于要快速查询，所以维护 $p_i$ 表示块内元素升序排序后的结果，查询时整块直接 lower_bound，散块暴力.
至于区间加，就在每个块上维护懒标记 $tag\textit{tag}$ ，整块直接加在 $tag\textit{tag}$ 上，散块直接暴力加然后重构块.
取 $B=nB=\sqrt{n}$ ，时间复杂度 $O(qnlog⁡n)O(q\sqrt{n\log n})$

Code

$O2)2.31\text{KB},536\text{ms},15.72\text{MB}\;\texttt{(in total, C++20 with O2)}$

#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;
}

struct Block {
	int n, size, blocks;
	vector<int> A, P, belong, L, R, tag;
	
	inline Block() {}
	inline Block(const vector<int>& a): A(a) {
		n = a.size();
		size = (int)sqrt(n);
		blocks = (n + size - 1) / size;
		
		P.resize(n);
		belong.resize(n);
		L.resize(blocks);
		R.resize(blocks);
		tag.resize(blocks);
		
		for (int i = 0; i < blocks; i++) {
			L[i] = i * size;
			R[i] = min(L[i] + size - 1, n - 1);
			for (int j = L[i]; j <= R[i]; j++) belong[j] = i;
		}
		for (int i = 0; i < blocks; i++) build(i);
	}
	
	inline void build(int b) {
		for (int i = L[b]; i <= R[b]; i++) {
			A[i] += tag[b];
			P[i] = A[i];
		}
		sort(P.begin() + L[b], P.begin() + R[b] + 1);
		tag[b] = 0;
	}
	
	inline void add(int l, int r, int x) {
		const int bl = belong[l], br = belong[r];
		if (bl == br) {
			for (int i = l; i <= r; i++) A[i] += x;
			return build(bl);
		}
		for (int i = l; i <= R[bl]; i++) A[i] += x;
		for (int i = L[br]; i <= r; i++) A[i] += x;
		build(bl), build(br);
		for (int i = bl + 1; i < br; i++) tag[i] += x;
	}
	
	inline int query(int l, int r, int x) {
		const int bl = belong[l], br = belong[r];
		if (bl == br) {
			int ans = 0;
			for (int i = l; i <= r; i++) ans += (A[i] + tag[bl] >= x);
			return ans;
		}
		
		int ans = 0;
		for (int i = l; i <= R[bl]; i++) ans += (A[i] + tag[bl] >= x);
		for (int i = L[br]; i <= r; i++) ans += (A[i] + tag[br] >= x);
		for (int i = bl + 1; i < br; i++) {
			ans += R[i] + 1 - (lower_bound(P.begin() + L[i], P.begin() + R[i] + 1, 
			                               x - tag[i]) - P.begin());
		}
		return ans;
	}
};

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int n, m;
	cin >> n >> m;
	
	vector<int> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	
	Block blk(a);
	for (int i = 0; i < m; i++) {
		char op; int l, r, x;
		cin >> op >> l >> r >> x;
		l--, r--;
		if (op == 'M') blk.add(l, r, x);
		else cout << blk.query(l, r, x) << endl;
	}
	
	return 0;
}