Open the Lock
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3400 Accepted Submission(s): 1507
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 搜索....bfs 这道题很经典.. 规则为 +1 ，-1，两个对调，若 9+1=1,1-1=9，同时最右边不能和最左边数字不能相邻..比如 3***4，就不行 单项搜索 算法..
1 #include<iostream>
2 #include<queue>
3 #include<set>
4 using namespace std;
5 typedef struct
6 {
7 int data;
8 int step;
9 }go;
10 int dir[4]={1,10,100,1000};
11 void bfs(int st,int en)
12 {
13 int i;
14 queue<go>mat;
15 set<int>visit; //用来存储是否产生这个数
16 go q,tem;
17 q.data=st;
18 q.step=0;
19 mat.push(q);
20 visit.insert(st);
21 while(!mat.empty())
22 {
23 tem=mat.front();
24 mat.pop();
25 if(tem.data==en)
26 {
27 printf("%d\n",tem.step);
28 return ;
29 }
30 /*先进行+1oper*/
31 for(i=0 ;i<4;i++)
32 {
33 q=tem;
34 if((q.data/dir[i])%10==9) q.data-=8*dir[i];
35 else
36 q.data+=dir[i];
37 q.step++;
38 if(visit.find(q.data)==visit.end()) //说明该状态没有
39 {
40 visit.insert(q.data);
41 mat.push(q);
42 }
43 }
44 /*进行-1 oper*/
45 for(i=0; i<4;i++)
46 {
47 q=tem ;
48 if((q.data/dir[i])%10==1) q.data+=8*dir[i];
49 else q.data-=dir[i];
50 q.step++;
51 if(visit.find(q.data)==visit.end()) //说明该状态没有
52 {
53 visit.insert(q.data);
54 mat.push(q);
55 }
56 }
57 /*对调旋转*/
58 int aa,bb;
59 for(i=0,q=tem;i<3;i++)
60 {
61 q=tem;
62 aa=(q.data/dir[i])%10;
63 bb=(q.data/dir[i+1])%10;
64 q.data=(q.data+(bb-aa)*dir[i]+(aa-bb)*dir[i+1]);
65 q.step++;
66 if(visit.find(q.data)==visit.end()) //说明该状态没有
67 {
68 visit.insert(q.data);
69 mat.push(q);
70 }
71 }
72 }
73 };
74
75 int main()
76 {
77 int st,en,t;
78 cin>>t;
79 while(t--)
80 {
81 scanf("%d%d",&st,&en);
82 bfs(st,en);
83 }
84 return 0;
85 }
View Code
采用双向广度搜索..
其实所谓双向广度，就是对于两边，每一次扩展下一层，就去扫一下，看对面有没有与之匹配的，状态
代码：
1 #include<cstdio>
2 #include<iostream>
3 #include<queue>
4 #include<set>
5 using namespace std;
6
7 typedef struct
8 {
9 int data;
10 int step;
11 /* void clr(){
12 step=0;
13
14 }*/
15 }go;
16 const int dir[4]={1,10,100,1000};
17 void Dbfs(int const st ,int const en)
18 {
19 go tem1,tem2,q;
20 tem1.data=st;
21 tem2.data=en;
22 /* tem1.clr();
23 tem2.clr();
24 */
25 tem1.step=tem2.step=0;
26 set<int>sav1,sav2;
27 sav1.insert(st);
28 sav2.insert(en);
29 queue<go> beg,end;
30 int i,a,b,cnt1,cnt2;
31 beg.push(tem1);
32 end.push(tem2);
33 cnt1=cnt2=0;
34 while(!beg.empty()&&!end.empty())
35 {
36 //+1oper
37 while(!beg.empty()&&cnt1==beg.front().step)
38 {
39 q=beg.front();
40 beg.pop();
41 if(sav2.find(q.data)!=sav2.end())
42 {
43 //说明有交集
44 while(end.front().data!=q.data)
45 end.pop();
46 printf("%d\n",q.step+end.front().step);
47 return ;
48 }
49 for(i=0;i<4;i++)
50 {
51 tem1=q;
52 if((tem1.data/dir[i])%10==9) tem1.data-=8*dir[i];
53 else
54 tem1.data+=dir[i];
55 tem1.step++;
56 if(sav2.find(tem1.data)!=sav2.end())
57 {
58 //说明有交集
59 while(end.front().data!=tem1.data)
60 end.pop();
61 printf("%d\n",tem1.step+end.front().step);
62 return ;
63 }
64 if(sav1.find(tem1.data)==sav1.end()) //标记
65 {
66 sav1.insert(tem1.data);
67 beg.push(tem1);
68 }
69 }
70 //-1oper
71 for(i=0;i<4;i++)
72 {
73 tem1=q;
74 if((tem1.data/dir[i])%10==1) tem1.data+=8*dir[i];
75 else
76 tem1.data-=dir[i];
77 tem1.step++;
78 if(sav2.find(tem1.data)!=sav2.end())
79 {
80 //说明有交集
81 while(end.front().data!=tem1.data)
82 end.pop();
83 printf("%d\n",tem1.step+end.front().step);
84 return ;
85 }
86 if(sav1.find(tem1.data)==sav1.end()) //标记
87 {
88 sav1.insert(tem1.data);
89 beg.push(tem1);
90 }
91 }
92 //旋转
93 for(i=0;i<3;i++)
94 {
95 tem1=q;
96 a=(tem1.data/dir[i])%10;
97 b=(tem1.data/dir[i+1])%10;
98 tem1.data+=(a-b)*(dir[i+1]-dir[i]);
99 tem1.step++;
100 if(sav2.find(tem1.data)!=sav2.end())
101 {
102 //说明有交集
103 while(end.front().data!=tem1.data)
104 end.pop();
105 printf("%d\n",tem1.step+end.front().step);
106 return ;
107 }
108 if(sav1.find(tem1.data)==sav1.end()) //标记
109 {
110 sav1.insert(tem1.data);
111 beg.push(tem1);
112 }
113 }
114 }
115 cnt1++;
116 while(!end.empty()&&cnt2==end.front().step)
117 {
118 q=end.front();
119 end.pop();
120 //+1oper
121 for(i=0;i<4;i++)
122 {
123 tem2=q;
124 if((tem2.data/dir[i])%10==9) tem2.data-=8*dir[i];
125 else
126 tem2.data+=dir[i];
127 tem2.step++;
128 if(sav1.find(tem2.data)!=sav1.end())
129 {
130 //说明有交集
131 while(beg.front().data!=tem2.data)
132 beg.pop();
133 printf("%d\n",tem2.step+beg.front().step);
134 return ;
135 }
136 if(sav2.find(tem2.data)==sav2.end()) //标记
137 {
138 sav2.insert(tem2.data);
139 end.push(tem2);
140 }
141 }
142 //-1oper
143 for(i=0;i<4;i++)
144 {
145 tem2=q;
146 if((tem2.data/dir[i])%10==1) tem2.data+=8*dir[i];
147 else
148 tem2.data-=dir[i];
149 tem2.step++;
150
151 if(sav1.find(tem2.data)!=sav1.end())
152 {
153 //说明有交集
154 while(beg.front().data!=tem2.data)
155 beg.pop();
156 printf("%d\n",tem2.step+beg.front().step);
157 return ;
158 }
159 if(sav2.find(tem2.data)==sav2.end()) //标记
160 {
161 sav2.insert(tem2.data);
162 end.push(tem2);
163 }
164 }
165 //旋转
166 for(i=0;i<3;i++)
167 {
168 tem2=q;
169 a=(tem2.data/dir[i])%10;
170 b=(tem2.data/dir[i+1])%10;
171 tem2.data+=(a-b)*(dir[i+1]-dir[i]);
172 tem2.step++;
173 if(sav1.find(tem2.data)!=sav1.end())
174 {
175 //说明有交集
176 while(beg.front().data!=tem2.data)
177 beg.pop();
178 printf("%d\n",tem2.step+beg.front().step);
179 return ;
180 }
181 if(sav2.find(tem2.data)==sav2.end()) //标记
182 {
183 sav2.insert(tem2.data);
184 end.push(tem2);
185 }
186 }
187 }
188 cnt2++;
189 }
190 }
191
192 int main()
193 {
194 int st,en,t;
195 scanf("%d",&t);
196 while(t--)
197 {
198 scanf("%d%d",&st,&en);
199 Dbfs( st , en );
200 }
201 return 0;
202 }
View Code
编程是一种快乐，享受代码带给我的乐趣！！！
