本文章由 WyOJ Shojo 从洛谷专栏拉取，原发布时间为 2022-07-27 11:02:03

题解思路：

我们对于 $c$ 相同的看成一个数。 那么我们倒着扫描，那么只要能放就放就可以了。

AC CODE：

#include <bits\/stdc++.h>
using namespace std;
#define ll long long
const int N = 2e5 + 5;
int n, k, cnt;
int a[N], b[N], p[N], fp[N], num[N];
vector<int> zz[N];
inline void init()
{
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; ++i)
	{
		int x;
		cin >> x;
		++num[x];
	}
	for (int i = 1; i <= k; ++i)
	{
		int x;
		cin >> x;
		if (!p[x])
			p[x] = ++cnt, fp[cnt] = x;
		a[cnt] += num[i];
		while (num[i]--)
			zz[cnt].push_back(i);
	}
	for (int i = 1; i <= n; ++i)
		b[i] = fp[i] - fp[i + 1];
	int x = 0;
	for (int i = cnt; i >= 1; --i)
	{
		if (!a[i])
			x += b[i];
		else
			b[i] += x, x = 0;
	}
}
int main()
{
	init();
	vector<vector<int>> ans;
	vector<pair<int, int>> now, last;
	for (int i = cnt; i >= 1; --i)
	{
		if (!a[i])
			continue;
		last.push_back({i, a[i]});
	}
	while (!last.empty())
	{
		int val = 0, z = 0;
		vector<int> res;
		for (auto &x : last)
		{
			int id = x.first, w = x.second;
			val += b[id];
			b[id] += z;
			int mi = min(x.second, val);
			w -= mi;
			val -= mi;
			while (mi--)
				res.push_back(id);
			if (w)
				now.push_back({id, w}), z = 0;
			else
				z = b[id];
		}
		ans.push_back(res);
		last = now;
		now.clear();
	}
	cout << ans.size() << '\n';
	for (auto &x : ans)
	{
		cout << x.size() << ' ';
		for (auto y : x)
		{
			cout << zz[y].back() << ' ';
			zz[y].pop_back();
		}
		cout << '\n';
	}
	return 0;
}