题目连接

Open the Lock

Description

Now an emergent task for you is to open a password lock. The password is consisted of four digits. Each digit is numbered from 1 to 9.

Each time, you can add or minus 1 to any digit. When add 1 to '9', the digit will change to be '1' and when minus 1 to '1', the digit will change to be '9'. You can also exchange the digit with its neighbor. Each action will take one step.

Now your task is to use minimal steps to open the lock.

Note: The leftmost digit is not the neighbor of the rightmost digit.

Input

The input file begins with an integer T, indicating the number of test cases.

Each test case begins with a four digit N, indicating the initial state of the password lock. Then followed a line with anotther four dight M, indicating the password which can open the lock. There is one blank line after each test case.

Output

For each test case, print the minimal steps in one line.

Sample Input

1234

2144

1111

9999

Sample Output

简单的广索题。。

1 #include
      
         2 #include
       
          3 #include
        
           4 #include
         
            5 #include
          
             6 #include
           
             7 #include
            
              8 #include
              9 using std::cin; 10 using std::cout; 11 using std::endl; 12 using std::swap; 13 using std::sort; 14 using std::map; 15 using std::pair; 16 using std::queue; 17 using std::vector; 18 using std::multimap; 19 #define pb(e) push_back(e) 20 #define sz(c) (int)(c).size() 21 #define mp(a, b) make_pair(a, b) 22 #define all(c) (c).begin(), (c).end() 23 #define iter(c) decltype((c).begin()) 24 #define cls(arr,val) memset(arr,val,sizeof(arr)) 25 #define cpresent(c, e) (find(all(c), (e)) != (c).end()) 26 #define rep(i, n) for (int i = 0; i < (int)(n); i++) 27 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i) 28 const int N = 10010; 29 typedef unsigned long long ull; 30 bool vis[N]; 31 int start, end; 32 struct Node { 33 int v, s; 34 Node(int i = 0, int j = 0) :v(i), s(j) {} 35 }; 36 inline void calc_1(int *arr, int v) { 37 int i, t, j = 0; 38 do arr[j++] = v % 10; while (v /= 10); 39 for (i = 0, --j; i < j; i++, j--) { 40 t = arr[i]; 41 arr[i] = arr[j]; 42 arr[j] = t; 43 } 44 } 45 inline int calc_2(int *arr) { 46 int sum = 0; 47 rep(i, 4) sum = sum * 10 + arr[i]; 48 return sum; 49 } 50 inline void calc_3(queue
              
                &q, int k, int s) { 51 if (!vis[k]) { 52 q.push(Node(k, s + 1)); 53 vis[k] = true; 54 } 55 } 56 int bfs() { 57 int v, k, tmp[10]; 58 cls(vis, false); 59 queue
               
                 q; 60 q.push(Node(start, 0)); 61 vis[start] = true; 62 while (!q.empty()) { 63 Node t = q.front(); q.pop(); 64 if (t.v == end) return t.s; 65 calc_1(tmp, t.v); 66 rep(i, 4) { 67 v = tmp[i]; 68 if (v + 1 == 10) tmp[i] = 1; 69 else tmp[i] = v + 1; 70 k = calc_2(tmp); 71 calc_3(q, k, t.s); 72 tmp[i] = v; 73 if (v - 1 == 0) tmp[i] = 9; 74 else tmp[i] = v - 1; 75 k = calc_2(tmp); 76 calc_3(q, k, t.s); 77 tmp[i] = v; 78 } 79 rep(i, 3) { 80 calc_1(tmp, t.v); 81 swap(tmp[i], tmp[i + 1]); 82 k = calc_2(tmp); 83 calc_3(q, k, t.s); 84 } 85 } 86 return 0; 87 } 88 int main() { 89 #ifdef LOCAL 90 freopen("in.txt", "r", stdin); 91 freopen("out.txt", "w+", stdout); 92 #endif 93 int t; 94 scanf("%d", &t); 95 while (t--) { 96 scanf("%d %d", &start, &end); 97 printf("%d\n", bfs()); 98 } 99 return 0;100 }

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转载于:https://www.cnblogs.com/GadyPu/p/4635111.html

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