Before Lunar New Year 2016 ends...

Let us analyse this circuit piece by piece. Due to the break in the circuit in the middle of the diagram, there are two separate circuits running from $A$ to $B$ . The top one consists of $2\Omega$ and $1\Omega$ in parallel, followed by $1\Omega$ in series, for an effective resistance of $\big(\tfrac11 + \tfrac12\big)^{-1} + 1 = \tfrac53\Omega$ . See the first row in the diagram above.

The grouping of six $1\Omega$ resistors is equivalent, after some vertex collapse, to a series of two triples of $1\Omega$ resistors in parallel, and hence has effective resistance $2\big(\tfrac11+ \tfrac11 + \tfrac11\big)^{-1} = \tfrac23\Omega$ . See the second row in the diagram.

The bottom circuit from $A$ to $B$ can thus be represented by the circuit in the bottom-left of the diagram. Suppose that there is a current $I$ passing through this circuit, and that there are circulant currents $I_1$ and $I_2$ in the two loops as indicated. Let the voltages at the four key vertices be $V$ , $V_1$ , $V_2$ and $V_3$ as indicated. Then the equations $\begin{array}{rclcrcl} V - V_1 & = & I_1 & \qquad \qquad & V - V_2 & = & I - I_1 \\ V_1 - V_2 & = & I_1 - I_2 & \qquad \qquad & V_1 - V_3 & = & I_2 \\ V_2 - V_3 & = & \tfrac23(I - I_2) \end{array}$ describe the potential losses across each component of the circuit. Solving these simultaneous equations gives $\begin{array}{rclcrclcrcl} I_1 & = & \tfrac{10}{21}I & \; & I_2 & = & \tfrac37I \\ V_1 & =& V - \tfrac{10}{21}I &\;& V_2 & = & V - \tfrac{11}{21}I &\;& V_3 & = & V - \tfrac{19}{21}I \end{array}$ Thus the effective resistance of the lower circuit is $\tfrac{V-V_3}{I} = \tfrac{19}{21}\Omega$ .

Thus the whole circuit is equivalent to resistors of $\tfrac53\Omega$ and $\tfrac{19}{21}\Omega$ in parallel, for an overall effective resistance of $\big(\tfrac35 + \tfrac{21}{19}\big)^{-1} = \tfrac{95}{162}\Omega$ . This makes the answer $95 + 162 = \boxed{257}$ .