HDUOJ---1195Open the Lock
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
搜尋....bfs 這道題很經典..
規則為 +1 ,-1,兩個對調,若 9+1=1,1-1=9,同時最右邊不能和最左邊數字不能相鄰..比如 3***4,就不行
單項搜尋
演算法..
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; //用來儲存是否產生這個數 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("%dn",tem.step); 28 return ; 29 } 30 /*先進行+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()) //說明該狀態沒有 39 { 40 visit.insert(q.data); 41 mat.push(q); 42 } 43 } 44 /*進行-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()) //說明該狀態沒有 52 { 53 visit.insert(q.data); 54 mat.push(q); 55 } 56 } 57 /*對調旋轉*/ 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()) //說明該狀態沒有 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 }
採用雙向廣度搜索..
其實所謂雙向廣度,就是對於兩邊,每一次擴充套件下一層,就去掃一下,看對面有沒有與之匹配的,狀態
程式碼:
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 //說明有交集
44 while(end.front().data!=q.data)
45 end.pop();
46 printf("%dn",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 //說明有交集
59 while(end.front().data!=tem1.data)
60 end.pop();
61 printf("%dn",tem1.step+end.front().step);
62 return ;
63 }
64 if(sav1.find(tem1.data)==sav1.end()) //標記
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 //說明有交集
81 while(end.front().data!=tem1.data)
82 end.pop();
83 printf("%dn",tem1.step+end.front().step);
84 return ;
85 }
86 if(sav1.find(tem1.data)==sav1.end()) //標記
87 {
88 sav1.insert(tem1.data);
89 beg.push(tem1);
90 }
91 }
92 //旋轉
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 //說明有交集
103 while(end.front().data!=tem1.data)
104 end.pop();
105 printf("%dn",tem1.step+end.front().step);
106 return ;
107 }
108 if(sav1.find(tem1.data)==sav1.end()) //標記
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 //說明有交集
131 while(beg.front().data!=tem2.data)
132 beg.pop();
133 printf("%dn",tem2.step+beg.front().step);
134 return ;
135 }
136 if(sav2.find(tem2.data)==sav2.end()) //標記
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 //說明有交集
154 while(beg.front().data!=tem2.data)
155 beg.pop();
156 printf("%dn",tem2.step+beg.front().step);
157 return ;
158 }
159 if(sav2.find(tem2.data)==sav2.end()) //標記
160 {
161 sav2.insert(tem2.data);
162 end.push(tem2);
163 }
164 }
165 //旋轉
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 //說明有交集
176 while(beg.front().data!=tem2.data)
177 beg.pop();
178 printf("%dn",tem2.step+beg.front().step);
179 return ;
180 }
181 if(sav2.find(tem2.data)==sav2.end()) //標記
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 }