Luogu3380 二逼平衡树

题目蓝链

Description

您需要写一种数据结构，来维护一个有序数列，其中需要提供以下操作：

1. 查询\(k\)在区间内的排名
2. 查询区间内排名为\(k\)的值
3. 修改某一位值上的数值
4. 查询\(k\)在区间内的前驱(前驱定义为严格小于x，且最大的数，若不存在输出-2147483647)
5. 查询\(k\)在区间内的后继(后继定义为严格大于x，且最小的数，若不存在输出2147483647)

Solution

首先我们可以想到一个非常暴力的做法，就是在普通平衡树上套一个线段树。这种做法除了二操作的复杂度是\(\mathcal{O}(\log^3 n)\)，其余均为\(\mathcal{O}(\log^2 n)\)。正是因为这样，所以这种做法用pbds代替平衡树之后，再开个O2，跑得比下面那种做法还快 早晓得这个跑这么快就打这个了

我们其实可以把内层改为权值线段树，然后外层再套一个树状数组/线段树(树状数组可以省一点点空间，线段树的常数是树状数组常数的一半) 我tm还不如直接写线段树。这样我们二操作就可以省去二分答案的那个\(\log\)，实现了所有操作复杂度均为\(\mathcal{O}(\log^2 n)\)(如果你不想离散化，还会乘上一个常数)

Code

#include <bits/stdc++.h>

using namespace std;

#define fst first
#define snd second
#define mp make_pair
#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef pair<int, int> pii;

template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }

inline int read() {
	int sum = 0, fg = 1; char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1;
	for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
	return fg * sum;
}

const int maxn = 5e4 + 10;
const int inf = 1e8;

pii operator + (const pii &a, const pii &b) { return mp(a.fst + b.fst, a.snd + b.snd); }

struct node {
	int ls, rs, v;
}A[maxn * 444];

struct sts {
	int a[16], cnt;
	sts(): cnt(0) { }
	void ins(int x) { a[++cnt] = x; }
	sts ls() {
		sts res; res.cnt = cnt;
		for (int i = 1; i <= cnt; i++) res.a[i] = A[a[i]].ls;
		return res;
	}
	sts rs() {
		sts res; res.cnt = cnt;
		for (int i = 1; i <= cnt; i++) res.a[i] = A[a[i]].rs;
		return res;
	}
	int sum() {
		int res = 0;
		for (int i = 1; i <= cnt; i++) res += A[a[i]].v;
		return res;
	}
};

int rt[maxn];
namespace ST {
	int cnt;
	void change(int &rt, int l, int r, int x, int v) {
		if (!rt) rt = ++cnt; A[rt].v += v;
		if (l == r) return;
		int mid = (l + r) >> 1;
		if (x <= mid) change(A[rt].ls, l, mid, x, v);
		else change(A[rt].rs, mid + 1, r, x, v);
	}
	pii query(sts a, sts b, int l, int r, int x) {
		if (l == r) return mp(0, b.sum() - a.sum());
		int mid = (l + r) >> 1;
		int tot = (b.ls()).sum() - (a.ls()).sum();
		if (x <= mid) return query(a.ls(), b.ls(), l, mid, x);
		return mp(tot, 0) + query(a.rs(), b.rs(), mid + 1, r, x);
	}
	int rank(sts a, sts b, int l, int r, int k) {
		if (l == r) return l;
		int mid = (l + r) >> 1;
		int tot = (b.ls()).sum() - (a.ls()).sum();
		if (k <= tot) return rank(a.ls(), b.ls(), l, mid, k);
		return rank(a.rs(), b.rs(), mid + 1, r, k - tot);
	}
}

int n, m, a[maxn];

namespace BIT {
#define lowbit(x) ((x) & -(x))
	void change(int x, int v, int op) {
		for (int i = x; i <= n; i += lowbit(i))
			ST::change(rt[i], 0, inf, v, op);
	}
	pii query(int l, int r, int k) {
		sts L, R;
		for (int i = r; i; i -= lowbit(i)) R.ins(rt[i]);
		for (int i = l - 1; i; i -= lowbit(i)) L.ins(rt[i]);
		return ST::query(L, R, 0, inf, k);
	}
	int rank(int l, int r, int k) {
		sts L, R;
		for (int i = r; i; i -= lowbit(i)) R.ins(rt[i]);
		for (int i = l - 1; i; i -= lowbit(i)) L.ins(rt[i]);
		return ST::rank(L, R, 0, inf, k);
	}
}

int main() {
#ifdef xunzhen
	freopen("tree.in", "r", stdin);
	freopen("tree.out", "w", stdout);
#endif

	n = read(), m = read();
	for (int i = 1; i <= n; i++) BIT::change(i, a[i] = read(), 1);

	for (int i = 1; i <= m; i++) {
		int op = read();
		if (op == 1) {
			int l = read(), r = read(), k = read();
			printf("%d\n", BIT::query(l, r, k).fst + 1);
		}
		if (op == 2) {
			int l = read(), r = read(), k = read();
			printf("%d\n", BIT::rank(l, r, k));
		}
		if (op == 3) {
			int x = read();
			BIT::change(x, a[x], -1), BIT::change(x, a[x] = read(), 1);
		}
		if (op == 4) {
			int l = read(), r = read(), x = read();
			pii res = BIT::query(l, r, x);
			if (res.fst) printf("%d\n", BIT::rank(l, r, res.fst));
			else printf("-2147483647\n");
		}
		if (op == 5) {
			int l = read(), r = read(), x = read();
			pii res = BIT::query(l, r, x);
			if (res.fst + res.snd < (r - l + 1)) printf("%d\n", BIT::rank(l, r, res.fst + res.snd + 1));
			else printf("2147483647\n");
		}
	}

	return 0;
}