Solution: As shown in the figure, make a symmetrical point a', connect a'b, cross the river at E, and pass B at F for BF⊥AC.

As can be seen from the figure, the quadrilateral CDBF is rectangular, so CD=FB=800m and DB = CF = 200 m.

Because the symmetry point of A is A', A'C = AC = 400m, AF = A' C+CF = 600 m..

∵BF⊥AC

∴∠A'FB=90

∫in△A ' FB，∠ a' FB = 90。

∴A'F^2+FB^2=AB^2

∴AB= 1000m

The shortest distance is 1000 meters.