HDU 1195 Open the Lock(BFS)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1195

 

Open the Lock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2587    Accepted Submission(s): 1140


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.
 

 

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 9999
 

 

Sample Output
2 4
 

 

Author
YE, Kai
 

 

Source
Zhejiang University Local Contest 2005
 

 

Recommend
Ignatius.L
 
 
BFS可以水过。。。。PE四次，没看清题，囧......据说双向BFS更快，可惜不会。。。。
 
贴下渣代码，等以后有机会改进下：
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<queue>
 5 #include<cmath>
 6 #include<cstdlib>
 7 using namespace std;
 8 int vis[10000],end;
 9 char start[4];
10 struct node
11 {
12     char num[5];
13     int time;
14 };
15 int convert(char s[])
16 {
17     int i,j,sum=0;
18     for(i=0,j=strlen(s)-1;i<strlen(s);i++,j--)
19     sum+=(s[i]-48)*pow(10.0,j*1.0);
20     return sum;
21 }
22 void bfs()
23 {
24     queue<node>q;
25     node p,s;
26     int i,temp;
27     char sh;
28     memset(vis,0,sizeof(vis));
29     p.time=0;
30     strcpy(p.num,start);
31     q.push(p);
32     while(!q.empty())
33     {
34         p=q.front();
35         q.pop();
36         if(convert(p.num)==end)//strcpy(p.num,end)
37         {
38             printf("%d\n",p.time);
39             return ;
40         }
41         for(i=0;i<4;i++)
42         {
43             strcpy(s.num,p.num);
44             if(s.num[i]<'9'&&s.num[i]>='1')s.num[i]+=1;
45             else if(s.num[i]=='9')s.num[i]='1';
46             s.num[4]='\0';
47             temp=convert(s.num);
48             if(!vis[temp])
49             {
50                 vis[temp]=1;
51                 s.time=p.time+1;
52                 q.push(s);
53             }
54             strcpy(s.num,p.num);
55             if(s.num[i]<='9'&&s.num[i]>'1')s.num[i]-=1;
56             else if(s.num[i]=='1')s.num[i]='9';
57             s.num[4]='\0';
58             temp=convert(s.num);
59             if(!vis[temp])
60             {
61                 vis[temp]=1;
62                 s.time=p.time+1;
63                 q.push(s);
64             }
65             if(i!=3)
66             {
67                 strcpy(s.num,p.num);
68                 sh=s.num[i];
69                 s.num[i]=s.num[i+1];
70                 s.num[i+1]=sh;
71                 s.num[4]='\0';
72                 temp=convert(s.num);
73                 if(!vis[temp])
74                 {
75                     vis[temp]=1;
76                     s.time=p.time+1;
77                     q.push(s);
78                 }
79             }
80         }
81     }
82 }
83 int main()
84 {
85     int t;
86     cin>>t;
87     while(t--)
88     {
89         cin>>start>>end;
90         getchar();
91         bfs();
92     }
93     return 0;
94 }

 

posted @ 2013-04-27 21:52  LFENG  阅读(220)  评论(0)  编辑  收藏  举报