hdu 1195:Open the Lock（暴力BFS广搜）
Open the Lock
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3847 Accepted Submission(s): 1661
Problem 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.
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.
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
9999
Sample Output
2
4
Author
YE, Kai
Source
Recommend
暴力BFS广搜。
利用广搜穷举所有情况。注意每一位可+1可-1，一共有4位，这就是8种情况，另外相邻两位可以交换位置，这又是3种情况。所以每一个节点一共要处理11种情况。用一个isv[][][][]四维数组记录可能出现的数字串出现过没有。例如，"2143"出现过了，则 isv[2][1][4][3]=true。
代码：
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include <queue>
5 using namespace std;
6 struct Node{
7 char a[4];
8 int step;
9 };
10 char a[5],b[5];
11 bool isv[10][10][10][10];
12 int bfs()
13 {
14 memset(isv,0,sizeof(isv));
15 queue <Node> q;
16 Node cur,next;
17 int i;
18 strcpy(cur.a,a);
19 cur.step = 0;
20 isv[cur.a[0]-'0'][cur.a[1]-'0'][cur.a[2]-'0'][cur.a[3]-'0'] = true;
21 q.push(cur);
22 while(!q.empty()){
23 cur = q.front();
24 q.pop();
25 //printf("%d\t%s\n",cur.step,cur.a);
26 for(i=0;i<4;i++)
27 if(cur.a[i] != b[i])
28 break;
29 if(i>=4) //找到
30 return cur.step;
31
32 //前八种，4个位置的数字+1或者-1
33 for(i=0;i<8;i++){
34 int nc = cur.a[i/2]-'0';
35 if(i%2) //+1
36 nc = nc%9+1;
37 else //-1
38 nc = nc==1?9:nc-1;
39 next = cur;
40 next.a[i/2] = nc + '0';
41 if(isv[next.a[0]-'0'][next.a[1]-'0'][next.a[2]-'0'][next.a[3]-'0'])
42 continue;
43 //可以放
44 isv[next.a[0]-'0'][next.a[1]-'0'][next.a[2]-'0'][next.a[3]-'0'] = true;
45 next.step = cur.step + 1;
46 q.push(next);
47 }
48 //后三种，相邻的数字交换位置
49 for(i=0;i<3;i++){
50 next = cur;
51 char t = next.a[i];
52 next.a[i] = next.a[i+1];
53 next.a[i+1] = t;
54 if(isv[next.a[0]-'0'][next.a[1]-'0'][next.a[2]-'0'][next.a[3]-'0'])
55 continue;
56 isv[next.a[0]-'0'][next.a[1]-'0'][next.a[2]-'0'][next.a[3]-'0'] = true;
57 next.step = cur.step + 1;
58 q.push(next);
59 }
60 }
61 return -1;
62 }
63 int main()
64 {
65 int i,T;
66 scanf("%d",&T);
67 for(i=1;i<=T;i++){
68 scanf("%s",a);
69 scanf("%s",b);
70 if(i!=1) //读取空行
71 scanf("%[^\n]%*");
72 int step = bfs();
73 printf("%d\n",step);
74 }
75 return 0;
76 }
Freecode : www.cnblogs.com/yym2013
