人工智能原理及其運用習題3.8

3.8  圖3.28是五個城市的交通圖,城市之間的連線旁邊的數字是城市之間路程的費用。要求從A城出發,通過其餘各城一次僅且一次,最後回到A城,請找出一條最優路線。spa

 

      解:五個城市能夠組成如下路徑:blog

            (1)A-B-C-D-E-A,代價=10+8+3+9+11=41.               (2)A-B-C-E-D-A,代價=10+8+8+9+9=44.im

            (3)A-B-D-C-E-A,代價=10+12+3+8+11=44.             (4)A-B-D-E-C-A,代價=10+12+9+8+2=41.img

            (5)A-B-E-C-D-A,代價=10+6+8+3+9=36.                 (6)A-B-E-D-C-A,代價=10+6+9+3+2=30.co

            (7)A-C-B-D-E-A,代價=2+8+12+9+11=42.               (8)A-C-B-E-D-A,代價=2+8+6+9+9=34.數字

            (9)A-C-D-B-E-A,代價=2+3+12+6+11=34.               (10)A-C-D-E-B-A,代價=2+3+9+6+10=30.ps

            (11)A-C-E-B-D-A,代價=2+8+6+12+9=37.               (12)A-C-E-D-B-A,代價=2+8+9+12+10=41.ab

            (13)A-D-B-C-E-A,代價=9+12+8+8+11=48.             (14)A-D-B-E-C-A,代價=9+12+6+8+2=37.

            (15)A-D-C-B-E-A,代價=9+3+8+6+11=37.               (16)A-D-C-E-B-A,代價=9+3+8+6+10=36.

            (17)A-D-E-C-B-A,代價=9+9+8+8+10=44.               (18)A-D-E-B-C-A,代價=9+9+6+8+2=34.

            (19)A-E-B-C-D-A,代價=11+6+8+3+9=37.               (20)A-E-B-D-C-A,代價=11+6+12+3+2=34.

            (21)A-E-C-B-D-A,代價=11+8+8+12+9=48.             (22)A-E-C-D-B-A,代價=11+8+3+12+10=44.

            (23)A-E-D-B-C-A,代價=11+9+12+8+2=42.             (24)A-E-D-C-B-A,代價=11+9+3+8+10=41.

             能夠看出最短路徑是  (6)A-B-E-D-C-A (10)A-C-D-E-B-A,他們是同一條路徑。

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