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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
2 1234 2144 1111 9999Sample Output
2 4
1 /* 2 By:OhYee 3 Github:OhYee 4 Email:oyohyee@oyohyee.com 5 Blog:http://www.cnblogs.com/ohyee/ 6 7 かしこいかわいい？ 8 エリーチカ！ 9 要写出来Хорошо的代码哦~ 10 */ 11 12 #include <cstdio> 13 #include <algorithm> 14 #include <cstring> 15 #include <cmath> 16 #include <string> 17 #include <iostream> 18 #include <vector> 19 #include <list> 20 #include <queue> 21 #include <stack> 22 #include <map> 23 using namespace std; 24 25 //DEBUG MODE 26 #define debug 0 27 28 //循环 29 #define REP(n) for(int o=0;o<n;o++) 30 31 int del(int a,int b) { 32 int cnt = 0; 33 while (a) { 34 int t1 = (a % 10); 35 int t2 = (b % 10); 36 cnt += abs(t1-t2)>4?(9 - abs(t1 - t2)): abs(t1 - t2); 37 a /= 10; 38 b /= 10; 39 } 40 return cnt; 41 } 42 43 int Swap(int s,int a,int b) { 44 int t[4] = {0}; 45 for (int i = 3;i >= 0;i--) { 46 t[i] = s % 10; 47 s /= 10; 48 } 49 swap(t[a],t[b]); 50 return t[0] * 1000 + t[1] * 100 + t[2] * 10 + t[3]; 51 } 52 53 int BFS(int s,int v) { 54 int Min = del(s,v); 55 bool visited[9999]; 56 memset(visited,false,sizeof(visited)); 57 58 queue<pair<int,int> > Q; 59 Q.push(pair<int,int>(s,0)); 60 visited[s] = true; 61 while (!Q.empty()) { 62 int k = Q.front().first; 63 int n = Q.front().second; 64 Q.pop(); 65 66 for (int i = 0;i < 3;i++) { 67 int kk = Swap(k,i,i + 1); 68 int nn = n + 1; 69 if (visited[kk]) 70 continue; 71 visited[kk] = true; 72 Min = min(nn + del(kk,v),Min); 73 //printf(" %d %d+%d=%d\n",kk,nn,del(kk,v),nn + del(kk,v)); 74 Q.push(pair<int,int>(kk,nn)); 75 } 76 } 77 return Min; 78 } 79 80 bool Do() { 81 int s,v; 82 scanf("%d%d",&s,&v); 83 84 printf("%d\n",BFS(s,v)); 85 86 return true; 87 } 88 89 int main() { 90 int T; 91 scanf("%d",&T); 92 while (T--) 93 Do(); 94 return 0; 95 }
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原文地址：http://www.cnblogs.com/ohyee/p/5428649.html
