[Solved]: Time complexity of solving a set of 2-sat instances?

Although 3SAT problem is NP-complete, there is a polynomial algorithm making use of the beautiful SCC (Strongly Connected Component) directed graph algorithm. Suppose there are $n$ variables and $m$ clauses in the 2SAT instance $mathcal{I}$. The parameters $n$ and $m$ represent the size of $mathcal{I}$ The 2-SAT instance is modeled by a directed graph $mathcal{G_I}$, with

• $2n$ vertices: for each variable $x$ in $mathcal{I}$, there are two vertices $v_x$ and $v_{bar{x}}$ (i.e., variable and its negation).
• $2m$ edges: for each clause $x lor y$, there are two directed edges $bar{x} to y$ and $bar{y} to x$. Note that the latter two implicative forms are equivalent to $x lor y$.

The polynomial algorithm (it is actually a linear algorithm) proceeds as follows:

1. Obtain the SCCs of $mathcal{G_{I}}$ in linear time $O(n + m)$;
2. Assign True to each literal (Notice: not the variable) in the destination SCC (denoted $SCC_d$);
3. Delete $SCC_d$ and its corresponding source SCC ($SCC_s$). (Note the symmetry in the graph $mathcal{G_{I}}$);
4. Repeat (2) and (3) until $mathcal{G_{I}}$ is empty.

For example: $mathcal{I} = (x_1 lor x_2) land (bar{x_2} lor x_3) land (bar{x_1} lor bar{x_2}) land (x_3 lor x_4) land (bar{x_3} lor x_5) land (bar{x_4} lor bar{x_5}) land (bar{x_3} lor x_4)$ The graph $mathcal{G_{I}}$ is: