= Solution
The two <complex scalar fields> have <scalar propagators>
$$
\langle T\phi\phi^\dagger\rangle(p)=\frac{i}{p^2-m^2+i\epsilon},\qquad
\langle T\Phi\Phi^\dagger\rangle(p)=\frac{i}{p^2-M^2+i\epsilon};
$$
contractions of two unconjugated fields or two conjugated fields vanish. Draw an arrow on each line in the direction of positive <conserved charge> flow: a solid line carries one unit and a dashed line two units. A particle arrow follows its physical <momentum>; an <antiparticle> arrow opposes it.
There are two conjugate three-leg <interaction vertices>, both with <Feynman rule> $-ig$: one has a charge-two arrow entering and two charge-one arrows leaving, and the other reverses every arrow. These describe the two conjugate terms $\phi^2\Phi^\dagger$ and $\phi^{\dagger2}\Phi$. The two identical $\phi$ attachments cancel the $1/2$ in the <Lagrangian density>. All crossed assignments are obtained by moving external lines between initial and final states while maintaining the arrows. There is no <interaction vertex> with three equal-sign charges.
Each <interaction vertex> conserves <four-momentum>, supplying $(2\pi)^4\delta^4(\sum p_j)$ with all momenta entering. Integrate each independent <loop momentum> with $d^4k/(2\pi)^4$ and divide by the <Feynman-diagram symmetry factor>. No minus sign is assigned to a closed scalar loop. These rules are for $i\mathcal M$ after amputating external lines and removing the overall <four-momentum conservation> factor.
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