STFT computational burden
The number of FFTs is roughly N/(w-m). An fft costs you nfft*log2(nfft) so the total is proportional to N/(w-m)*nfft*log2(nfft). 

A typical application would make nfft = w and m = w/2, so this would come out to be proportional to N*log2(nfft). So it's proportional to the length or you data set (obviously) and the log of the FFT size.