Multiset 题解

考虑求排名怎么做。

一种空间比较优秀的做法就是树状数组加二分找排名，即值域树状数组上二分。

然后删除即在树状数组上将数的值 $-1$ 即可。

复杂度应该是 $O(n \log n + q \log^2 n)$。常数有点大，加了快读卡一卡就过去了。

#pragma GCC optimize("-Ofast,inline,fast-math")
#pragma GCC target("avx,sse,sse2,sse3,sse4")
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;

const int N = 1e6 + 5;

inline int read()
{
	char ch(getchar());
	int x = 0, o = 1;
	while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
	while (ch == '-') o = -o, ch = getchar();
	while (ch >= '0' and ch <= '9')
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * o;
}

int n, a[N], q, k;

class Bit
{
public:
	int tr[N];
	void update(int u, int x)
	{
		while (u <= n)
		{
			tr[u] += x;
			u += u & -u;
		}
	}
	int query(int u)
	{
		int res = 0;
		while (u)
		{
			res += tr[u];
			u -= u & -u;
		}
		return res;
	}
}tr;

int main()
{
	n = read(), q = read();
	for (int i = 1; i <= n; ++i)
	{
		a[i] = read();
		tr.update(a[i], 1);
	}
	while (q--)
	{
		k = read();
		if (k < 0)
		{
			int l = 1, r = n, res = 0;
			while (l <= r)
			{
				int mid = l + r >> 1;

				if (tr.query(mid) >= -k)
				{
					r = mid - 1;
					res = mid;
				}
				else
				{
					l = mid + 1;
				}
			}
			if (res)
			{
				tr.update(res, -1);
			}
		}
		else
		{
			tr.update(k, 1);
		}
	}
	for (int i = 1; i <= n; ++i)
	{
		if (tr.query(i) - tr.query(i - 1))
		{
			printf("%d\n", i);
			return 0;
		}
	}
	printf("0\n");
	return 0;
}