Algorithm 3: Low-complexity subcarrier scheduling and user pairing scheme.
1:	Input:$N_c$, U, ${\left|l_{n_c,k_{n_c}}\right|}^2$, $P_{1,n_c}$ and $P_{2,n_c}$.
2:	Output: Subcarrier scheduling $M_{\mathrm{low}}$ and user pairing $k_{n_c}$.
3:	Initialize: $M_{\mathrm{low}}=\varnothing$, $i_1=1$ and $\varsigma =1$, where $M_{\mathrm{low}}$, i and $\varsigma$ denote the subcarrier mapping, ith element of ${\mathbb{U}}_s$ and $\varsigma$th user pairing result, respectively;
4:	Arrange the elements of ${\mathbb{U}}_s$ in descending order, known as $\overline{{\mathbb{U}}_s}$;
5:	while$i\in \left[1,⌈\frac{U}{2}⌉\right]$do
6:	Denote the subset of the use pairing as $\overline{{\mathbb{U}}_s^{\mathrm{sub}}}$;
7:	if $i+⌈\frac{U}{2}⌉=U$ then
8:	$\overline{{\mathbb{U}}_s^{\mathrm{sub}}}=\left[\left(\overline{{\mathbb{U}}_s}\left(i\right),\overline{{\mathbb{U}}_s}\left(i+⌈\frac{U}{2}⌉\right)\right),\left(\overline{{\mathbb{U}}_s}\left(i+⌈\frac{U}{2}⌉\right),\overline{{\mathbb{U}}_s}\left(i\right)\right)\right]$;
9:	else
10:	$\overline{{\mathbb{U}}_s^{\mathrm{sub}}}=\left[\left(\overline{{\mathbb{U}}_s}\left(i\right),\overline{{\mathbb{U}}_s}\left(i\right)\right),\left(\overline{{\mathbb{U}}_s}\left(i\right),\overline{{\mathbb{U}}_s}\left(i\right)\right)\right]$;
11:	end if
12:	while $\varsigma \in [1,2]$ do
13:	Select the $\varsigma$th element of $\overline{{\mathbb{U}}_s^{\mathrm{sub}}}$;
14:	Determine the 1th and 2th users;
15:	Set $\mathbb{C}=\left\{\widetilde{n_c}|\widetilde{n_c}\in M_{\mathrm{low}}\right\}$, $\widetilde{n_c}$ is the unsaturated points;
16:	if $\mathbb{C}\ne \varnothing$ then
17:	Find a new augmented branch $\chi$;
18:	$M_{\mathrm{low}}=M_{\mathrm{low}}⨁\chi$;
19:	else
20:	Calculate $E_{\mathrm{total}}$ as (16);
21:	Record $E_{\mathrm{total}}$, subcarrier scheduling $M_{\mathrm{low}}$ and user pairing $k_{n_c}$;
22:	$\varsigma =\varsigma +1$;
23:	end if
24:	$i=i+1$;
25:	end while
26:	$i=1$;
27:	end while
28:	Search the maximum $E_{\mathrm{total}}$ and corresponding subcarrier scheduling $M_{\mathrm{low}}$ and user pairing $k_{n_c}$.