P4396 [AHOI2013] 作业

题意

给定一个序列，每次询问求：

在区间 \([l, r]\) 中，大小在 \([a, b]\) 中数的个数与种类数。

Sol

对于第一问直接离线跑树状数组二维偏序。

第二问考虑莫队，发现只需要维护莫队那个表示种类的数组的区间和就行了。

要求 \(O(1)\) 修改的话，写个值域分块？

Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <tuple>
using namespace std;
#ifdef ONLINE_JUDGE

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;

#endif
int read() {
	int p = 0, flg = 1;
	char c = getchar();
	while (c < '0' || c > '9') {
		if (c == '-') flg = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9') {
		p = p * 10 + c - '0';
		c = getchar();
	}
	return p * flg;
}
void write(int x) {
	if (x < 0) {
		x = -x;
		putchar('-');
	}
	if (x > 9) {
		write(x / 10);
	}
	putchar(x % 10 + '0');
}

const int N = 1e5 + 5, M = 605;
array <int, N> s;

namespace Bit {

array <int, N> edge;

int lowbit(int x) {
	return x & -x;
}

void modify(int x, int y, int n) {
	if (!x) return;
	while (x <= n) {
		edge[x] += y;
		x += lowbit(x);
	}
	return;
}

int query(int x) {
	int ans = 0;
	while (x) {
		ans += edge[x];
		x -= lowbit(x);
	}
	return ans;
}

struct Node {
	int l, r, id, k;
	bool operator <(const Node &t) const {
		return l < t.l;
	}
};

}

namespace Cpt {

int bsi = 332;

struct Node {
	int l, r, x, y, id;
	bool operator < (const Node &t) const {
		if (l / bsi != t.l / bsi) return l < t.l;
		return (l / bsi) & 1 ? r < t.r : r > t.r;
	}
};

namespace Blk {

array <int, N> cur;
array <int, M> bk;

int calc(int x) {
	return (x - 1) / bsi + 1;
}

int query(int x, int y) {
	int ans = 0;
	if (calc(x) == calc(y)) {
		for (int i = x; i <= y; i++)
			ans += (cur[i] && 1);
		return ans;
	}
	for (int i = x; i <= calc(x) * bsi; i++)
		ans += (cur[i] && 1);
	for (int i = calc(x) + 1; i < calc(y); i++)
		ans += bk[i];
	for (int i = (calc(y) - 1) * bsi + 1; i <= y; i++)
		ans += (cur[i] && 1);
	return ans;
}

}

void add(int x) {
	Blk::cur[s[x]]++;
	if (Blk::cur[s[x]] == 1)
		Blk::bk[Blk::calc(s[x])]++;
}

void del(int x) {
	Blk::cur[s[x]]--;
	if (Blk::cur[s[x]] == 0)
		Blk::bk[Blk::calc(s[x])]--;
}

}

array <Bit::Node, N * 4> isl;
array <Cpt::Node, N> qrl;

array <int, N> ans1, ans2;

using Cpt::add; using Cpt::del;

int main() {
	int n = read(), m = read();
	for (int i = 1; i <= n; i++)
		s[i] = read();
	int cnt = 0;
	for (int i = 1; i <= m; i++) {
		int l = read(), r = read(), x = read(), y = read();
		qrl[i].l = l, qrl[i].r = r, qrl[i].x = x, qrl[i].y = y;
		qrl[i].id = i;
		cnt++; isl[cnt].l = r, isl[cnt].r = y, isl[cnt].id = i, isl[cnt].k = 1;
		cnt++; isl[cnt].l = l - 1, isl[cnt].r = y, isl[cnt].id = i, isl[cnt].k = -1;
		cnt++; isl[cnt].l = r, isl[cnt].r = x - 1, isl[cnt].id = i, isl[cnt].k = -1;
		cnt++; isl[cnt].l = l - 1, isl[cnt].r = x - 1, isl[cnt].id = i, isl[cnt].k = 1;
	}
	sort(isl.begin() + 1, isl.begin() + 1 + cnt);
	int tp = 0;
	for (int i = 0; i <= n; i++) {
		Bit::modify(s[i], 1, 100000);
		while (isl[tp + 1].l == i && tp + 1 <= cnt) {
			tp++;
			ans1[isl[tp].id] += isl[tp].k * Bit::query(isl[tp].r);
		}
	}
	/* for (int i = 1; i <= m; i++) */
		/* write(ans1[i]), putchar(32); */
	/* puts(""); */
	sort(qrl.begin() + 1, qrl.begin() + 1 + m);
	int l = 1, r = 0;
	for (int i = 1; i <= m; i++) {
		int x = qrl[i].l, y = qrl[i].r;
		while (l > x) l--, add(l);
		while (r < y) r++, add(r);
		while (l < x) del(l), l++;
		while (r > y) del(r), r--;
		ans2[qrl[i].id] = Cpt::Blk::query(qrl[i].x, qrl[i].y);
	}
	for (int i = 1; i <= m; i++) {
		write(ans1[i]), putchar(32);
		write(ans2[i]), puts("");
	}
	return 0;
}