CF940F Machine Learning

题意

给定 $n$ 个数的序列 $a$，有 $m$ 次操作，每次操作有单点修改和询问区间 $a_l, a_{l+1}, \cdots, a_r$ 中每个数的出现次数的 mex。

解法

考虑莫队维护。

我们要维护 mex，并且是带修莫队，所以回滚莫队用不了。我们考虑在获得答案时暴力。

考虑维护两个数组，分别是 $ca$ 和 $cb$，$ca$ 记录每个数出现次数，$cb$ 记录每个 $ca$ 的出现次数。

询问答案时从 $1$ 到 $\infty$ 暴力枚举，复杂度是可以接受的。

代码：

#pragma GCC optimize("-Ofast")
#pragma GCC target("avx,sse,sse2,sse3,sse4")
#pragma unroll 10
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;

const int N = 2e5 + 10;

int a[N], len, ans[N];
vector<int> b;

int n, m, cnta, cntb;

int ca[N], cb[N];

struct Query
{
	int id, l, r, k;
}q[N];

struct Modify
{
	int x, y;
}g[N];

inline int read()
{
	register char ch = getchar();
	register int x = 0;
	while (ch < '0' || ch > '9') ch = getchar();
	while (ch >= '0' && ch <= '9')
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x;
}

inline bool cmp(const Query& x, const Query& y)
{
	register int xl = x.l / len, xr = x.r / len;
	register int yl = y.l / len, yr = y.r / len;
	if (xl ^ yl) return xl < yl;
	if (xr ^ yr) return xr < yr;
	return x.k < y.k;
}

inline void add(int x)
{
	--cb[ca[a[x]]];
	++cb[++ca[a[x]]];
}

inline void del(int x)
{
	--cb[ca[a[x]]];
	++cb[--ca[a[x]]];
}

inline void change(int x, int l, int r)
{
	int rem = 0;
	if (g[x].x >= l && g[x].x <= r)
	{
		--cb[ca[a[g[x].x]]];
		++cb[--ca[a[g[x].x]]];
		--cb[ca[g[x].y]];
		++cb[++ca[g[x].y]];
	}
	swap(a[g[x].x], g[x].y);
}

int main()
{
	n = read();
	m = read();
	for (register int i = 1; i <= n; ++i)
	{
		a[i] = read();
		b.push_back(a[i]);
	}
	for (register int i = 1; i <= m; ++i)
	{
		register int opt = read(), l = read(), r = read();
		if (opt == 1)
		{
			++cnta;
			q[cnta] = { cnta, l, r, cntb };
		}
		else
		{
			++cntb;
			b.push_back(r);
			g[cntb] = { l, r };
		}
	}
	sort(b.begin(), b.end());
	b.erase(unique(b.begin(), b.end()), b.end());
	for (register int i = 1; i <= n; ++i)
	{
		a[i] = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
	}
	for (register int i = 1; i <= cntb; ++i)
	{
		g[i].y = lower_bound(b.begin(), b.end(), g[i].y) - b.begin();
	}
	len = pow(n, 2.0 / 3);
	sort(q + 1, q + cnta + 1, cmp);
	register int nl = 1, nr = 0, now = 0;
	for (register int i = 1; i <= cnta; ++i)
	{
		int l = q[i].l, r = q[i].r, k = q[i].k, ansk = 0;
		while (nl < l) del(nl++);
		while (nl > l) add(--nl);
		while (nr < r) add(++nr);
		while (nr > r) del(nr--);
		while (now < k) change(++now, l, r);
		while (now > k) change(now--, l, r);
		for (register int j = 1; ; ++j)
		{
			if (!cb[j])
			{
				ansk = j;
				break;
			}
		}
		ans[q[i].id] = ansk;
	}
	for (register int i = 1; i <= cnta; ++i) printf("%d\n", ans[i]);
	return 0;
}