【分块】【P2801】教主的魔法

Description

给你一个长度为 \(n\) 的序列，要求资瓷区间加，查询区间大于等于 \(k\) 的数的个数

Input

第一行是 \(n~,~Q\) 代表序列长度和操作个数

下面一行代表序列

下面 \(Q\) 行，每行四个参数，分别为 \(opt~,~l,~r~,w\)

如果 \(opt~=~M\) 则区间加

如果 \(opt~=~A\) 则查询

Output

对每次查询输出结果

Hint

\(1~\leq~n~\leq~1000000~,~1~\leq~q~\leq~1000\)

Solution

看到序列这么长，操作数这么少，常见的复杂度平衡的数据结构大概不起作用，于是考虑分块。

分块后考虑块内如何查询答案：可以对块内整个进行排序，然后lowerbound一下即可。

考虑修改时，如果修改整个块则不会对块内大小顺序造成影响，可以直接处理。

边界上直接暴力修改，修改完暴力排序，因为只会暴力两个块，所以复杂度还是 \(O(\sqrt{n}~(\log {\sqrt{n}})\) 的，总复杂度 \(O(q~\sqrt{n}~\log \sqrt{n})\)，可以通过本题。

Code

#include <cmath>

#include <cstdio>

#include <algorithm>

#ifdef ONLINE_JUDGE

#define freopen(a, b, c)

#endif

#define rg register

#define ci const int

#define cl const long long

typedef long long int ll;

namespace IPT {

	const int L = 1000000;

	char buf[L], *front=buf, *end=buf;

	char GetChar() {

		if (front == end) {

			end = buf + fread(front = buf, 1, L, stdin);

			if (front == end) return -1;

		}

		return *(front++);

	}

}

template <typename T>

inline void qr(T &x) {

	rg char ch = IPT::GetChar(), lst = ' ';

	while ((ch > '9') || (ch < '0')) lst = ch, ch=IPT::GetChar();

	while ((ch >= '0') && (ch <= '9')) x = (x << 1) + (x << 3) + (ch ^ 48), ch = IPT::GetChar();

	if (lst == '-') x = -x;

}

template <typename T>

inline void ReadDb(T &x) {

	rg char ch = IPT::GetChar(), lst = ' ';

	while ((ch > '9') || (ch < '0')) lst = ch, ch = IPT::GetChar();

	while ((ch >= '0') && (ch <= '9')) x = x * 10 + (ch ^ 48), ch = IPT::GetChar();

	if (ch == '.') {

		ch = IPT::GetChar();

		double base = 1;

		while ((ch >= '0') && (ch <= '9')) x += (ch ^ 48) * ((base *= 0.1)), ch = IPT::GetChar();

	}

	if (lst == '-') x = -x;

}

namespace OPT {

	char buf[120];

}

template <typename T>

inline void qw(T x, const char aft, const bool pt) {

	if (x < 0) {x = -x, putchar('-');}

	rg int top=0;

	do {OPT::buf[++top] = x % 10 + '0';} while (x /= 10);

	while (top) putchar(OPT::buf[top--]);

	if (pt) putchar(aft);

}

const int maxn = 1000010;

int n, q;

ll MU[maxn], belong[maxn], temp[maxn], tag[maxn], lc[maxn], rc[maxn];

void rebuild(ci);

int main() {

	freopen("1.in", "r", stdin);

	qr(n); qr(q);

	for (rg int i = 1; i <= n; ++i) qr(MU[i]);

	for (rg int i = 1; i <= n; ++i) temp[i] = MU[i];

	for (rg int i = 1, sn = sqrt(n); i <= n; ++i) belong[i] = i / sn;

	for (rg int i = 1, j = 1; i <= n; i = j) {

		while(belong[j] == belong[i]) ++j;

		std::sort(temp + i, temp + j);

		lc[belong[i]] = i; rc[belong[i]] = j;

	}

	int a, b, c; rg char ch;

	while (q--) {

		do {ch = IPT::GetChar();} while((ch != 'M') && (ch != 'A'));

		if (ch == 'M') {

			a = b = c = 0;

			qr(a); qr(b); qr(c);

			if (belong[a] == belong[b]) {

				for (rg int i = a; i <= b; ++i) {

					MU[i] += c;

				}

				rebuild(a);

			} else {

				for (rg int i = belong[a] + 1; i < belong[b]; ++i) tag[i] += c;

				for (rg int i = a; belong[i] == belong[a]; ++i) MU[i] += c;

				for (rg int i = b; belong[i] == belong[b]; --i) MU[i] += c;

				rebuild(a); rebuild(b);

			}

		} else {

			a = b = c = 0;

			qr(a); qr(b); qr(c);

			int _ret = 0;

			if (belong[a] == belong[b]) {

				c -= tag[belong[a]];

				for (rg int i = a; i <= b; ++i) if (MU[i] >= c) ++_ret;

			} else {

				for (rg int i = belong[a] + 1; i < belong[b]; ++i) _ret += temp + rc[i] - std::lower_bound(temp + lc[i], temp + rc[i], c - tag[i]);

				c -= tag[belong[a]];

				for (rg int i = a; belong[i] == belong[a]; ++i) _ret += MU[i] >= c;

				c += tag[belong[a]]; c -= tag[belong[b]];

				for (rg int i = b; belong[i] == belong[b]; --i)	_ret += MU[i] >= c;

			}

			qw(_ret, '\n', true);

		}

	}

	return 0;

}

void rebuild(ci a) {

	for (rg int i = lc[belong[a]]; belong[i] == belong[a]; ++i) temp[i] = MU[i];

	std::sort(temp + lc[belong[a]], temp + rc[belong[a]]);

}

Summary

当查询较少但是序列较长时，考虑分块来降低维护代价。

巴特西