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Efficient Path Finding Algorithm for Restricted Numbers

Find the shortest path from an initial number to a target number while avoiding certain forbidden numbers. Utilizes a BFS approach with 8 possible directions for each digit.

kyra-rogers
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Efficient Path Finding Algorithm for Restricted Numbers

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  1. 10067: Product ★★★★☆ 題組：Problem Set Archive with Online Judge 題號：10067: Product 解題者：雷皓博 解題日期：2014年5月5日 題意：給你初始值(ex 8056)和目地數字(ex 6508),並給你一些不可以到的數字,求到達目地數字的最少的步驟,無法到達的話輸出-1. (ex 8056 -> 8956 算一個步驟 ) (ex 8056 -> 8156 算一個步驟 ) 1

  2. 題意範例： INPUT 1 ->幾組測資 8 0 5 6 ->初始值 6 5 0 8 ->目地數字 5 ->幾個不可到的數字 8 0 5 7 ->可到的數字 8 0 4 7 5 5 0 8 7 5 0 8 6 4 0 8 OUTPUT 14 2

  3. 解法用BFS找最少步驟,每個數字有8個方向. (8056有8057,8055, 8066,8046, 8156,8956, 7056,9056) 我們可以把每一組4位數字都當成一個狀態,也就是0000~9999 共一萬組狀態,外加跑過的要刪掉 (可用一個array[10000]來記錄)

  4. 解法範例： array 先記錄不可到的數字. 用struct node 把 8056 跟num = 0包起來 Step1:剛開始先把初始值8056丟入queue裡 Step2: while(!queue.empty() ) 若queue是空得,跳到Step7 Step3:拿queue.front的值(current=queue.front),並pop Step4:判斷和目地數字是否相同,是的話跳出迴圈,並印出current的num Step5:把current 裡的數字的8個方向(array裡有的數字要除外)的值跟num+1都放入queue,並且array紀錄跑過的數字. Step6:到Step2繼續Loop. Step7: 都找不到則印出-1 討論：無

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