树状数组学习

树状数组

算法	单点修改	求前缀和
前缀和	\(O(1)\)	\(O(n)\)
树状数组	\(O(logn)\)	\(O(logn)\)

​ 我们将区间拆分成\([r - lowbit(r) + 1, r]\), 如上图所示，我们用右端点\(r\)来表示一个区间, 则\(r_1\)节点的父亲节点为\(r_1 + lowbit(r_1)\), 由于树状数组的每个节点到其祖宗节点是一条单链结构，所以对于单点修改，我们可以从儿子节点通过\(+lowbit(i)\) 枚举到父亲节点， 对于求前缀和操作我们只需要从\(x\)通过\(- lowbit(i)\) 依次向上枚举。

扩展

区间修改+单点查询

​ 由于树状数组可以\(O(logn)\)时间内完成求前缀操作，所以我们可以用树状数组维护一个差分数组, 求前缀即为第i个数的增加或减少值

区间修改+区间求和

​ 假设我们用树状数组来维护一个差分序列，则对于第i个数的数值，我们需要求

\[\sum \limits_{i = 1}^x \sum \limits_{j = 1}^i b_j \\ = (x + 1) \times (b_1 + b_2 + \dots + b_x) + (b_1 + 2b_2 + 3b_3 + \dots + xb_x) \\ = (x + 1) \sum \limits_{i = 1}^x b_i + \sum \limits_{i = 1}^xib_i \]

​ 通过上式，我们只需要用树状数组维护\(b_i\) 和\(ib_i\)的值即可， 通过求前缀操作便可以\(O(logn)\)时间内完成查询和修改操作

求逆序对

例题

一个简单的整数问题

#include <iostream>
#define lowbit(x) (x & -x)

const int N = 1e5 + 10;

using i64 = long long;

int n, m, a[N];
i64 tr[N];

void add(int x, i64 c) {
	for (int i = x; i <= n; i += lowbit(i)) tr[i] += c;
}

i64 ask(int x) {
	i64 res = a[x];
	for (int i = x; i; i -= lowbit(i)) res += tr[i];
	return res;
}

int main() {
	std::cin >> n >> m;

	for (int i = 1; i <= n; i ++) std::cin >> a[i];
	
	while (m --) {
		std::string op;
		int x, l, r;

		std::cin >> op;

		if (op == "Q") {
			std::cin >> x;
			std::cout << ask(x) << "\n";
		} else {
			std::cin >> l >> r >> x;
			add(l, x);
			add(r + 1, -x);
		}
	}
}

一个简单的整数问题2

#include <iostream>
#define lowbit(x) (x & -x)
using i64 = long long;

const int N = 1e5 + 10;

int n, m, a[N];

i64 tr1[N], tr2[N];

void add(i64 tr[], int x, i64 c) {
	for (int i = x; i <= n; i += lowbit(i)) tr[i] += c;
}

i64 query(i64 tr[], int x) {
	i64 res = 0;
	for (int i = x; i; i -= lowbit(i)) res += tr[i];
	return res;
}

i64 prefix(int x) {
	return query(tr1, x) * (x + 1) - query(tr2, x);
}

int main() {
	std::cin >> n >> m;

	for (int i = 1; i <= n; i ++) {
		std::cin >> a[i];
		add(tr1, i, a[i] - a[i - 1]);
		add(tr2, i, (i64)i * (a[i] - a[i - 1]));
	}

	while (m --) {
		char op[2];
		int l, r, x;

		scanf("%s%d%d", op, &l, &r);

		if (op[0] == 'C') {
			scanf("%d", &x);
			add(tr1, l, x), add(tr1, r + 1, -x);
			add(tr2, l, (i64) l * x), add(tr2, r + 1, (i64) (r + 1) * -x);
		} else {
			std::cout << prefix(r) - prefix(l - 1) << "\n";
		}
	}
}

谜一样的牛

#include <iostream>
#define lowbit(x) (x & -x)
using i64 = long long;

const int N = 1e5 + 10;

int n, tr[N], a[N], ans[N];

void add(int x, int c) {
	for (int i = x; i <= n; i += lowbit(i)) tr[i] += c;
}

int query(int x) {
	int res = 0;
	for (int i = x; i; i -= lowbit(i)) res += tr[i];
	return res;
}

int main() {
	std::cin >> n;

	for (int i = 2; i <= n; i ++) std::cin >> a[i];

	for (int i = 1; i <= n; i ++) tr[i] = lowbit(i);
	
	for (int i = n; i >= 1; i --) {
		int k = a[i] + 1;
		int l = 1, r = n;
		while (l < r) {
			int mid = l + r >> 1;
			if (query(mid) >= k) r = mid;
			else l = mid + 1;
		}
		add(r, -1);
		ans[i] = r;
	}

	for (int i = 1; i <= n; i ++) {
		std::cout << ans[i] << " \n"[i == n];
	}
}