There are two ways to know this problem from the meaning of the question, the shortest distance.

∵①A→CD→B, in which there are five methods for A → C. D → B has 1 methods, with a total of 5× 1=5 methods.

②A→EF→B, where A→C has 10 method (red 4 method, blue 3 method, green 2 method, yellow 1 method), and F→B has 3 methods, totaling 10×3=30 method.

∴ There are 5+30=35 short-distance routes from A to B in total.

So the answer is: 35.