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HDU 1195 Open the Lock
Open the Lock
Time Limit: 1000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 119564-bit integer IO format: %I64d Java class name: Main
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
21234214411119999
Sample Output
24
Source
Zhejiang University Local Contest 2005
解题：bfs...
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 struct node{18 int key[4],step;19 };20 queue<node>q;21 set<int>st;22 int temp[4],tar[4];23 bool transformIt(int *d){24 int num = 0;25 for(int i = 0; i < 4; i++)26 num = num*10 + d[i];27 if(st.count(num)) return false;28 st.insert(num);29 return true;30 }31 bool ok(int *d){32 for(int i = 0; i < 4; i++)33 if(d[i] != tar[i]) return false;34 return true;35 }36 int bfs(){37 while(!q.empty()) q.pop();38 node a,b;39 memcpy(a.key,temp,sizeof(temp));40 st.clear();41 a.step = 0;42 q.push(a);43 transformIt(a.key);44 while(!q.empty()){45 b = q.front();46 q.pop();47 if(ok(b.key)) return b.step;48 for(int i = 0; i < 4; i++){49 a.step = b.step+1;50 memcpy(temp,b.key,sizeof(temp));51 if(temp[i] + 1 > 9) temp[i] = 1;52 else temp[i]++;53 if(transformIt(temp)){54 memcpy(a.key,temp,sizeof(temp));55 q.push(a);56 }57 memcpy(temp,b.key,sizeof(temp));58 if(temp[i]-1 == 0) temp[i] = 9;59 else temp[i]--;60 if(transformIt(temp)){61 memcpy(a.key,temp,sizeof(temp));62 q.push(a);63 }64 if(i){65 memcpy(temp,b.key,sizeof(temp));66 swap(temp[i],temp[i-1]);67 if(transformIt(temp)){68 memcpy(a.key,temp,sizeof(temp));69 q.push(a);70 }71 }72 if(i < 3){73 memcpy(temp,b.key,sizeof(temp));74 swap(temp[i],temp[i+1]);75 if(transformIt(temp)){76 memcpy(a.key,temp,sizeof(temp));77 q.push(a);78 }79 }80 }81 }82 return -1;83 }84 int main(){85 int t,i,j;86 char md[6];87 scanf("%d",&t);88 while(t--){89 scanf("%s",md);90 for(i = 0; i < 4; i++) temp[i] = md[i] - ‘0‘;91 scanf("%s",md);92 for(i = 0; i < 4; i++) tar[i] = md[i] - ‘0‘;93 printf("%d\n",bfs());94 }95 return 0;96 }
不用set居然快了这么多。。。。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 struct node{18 int key[4],step;19 };20 queue<node>q;21 int temp[4],tar[4];22 bool vis[10000];23 bool transformIt(int *d){24 int num = 0;25 for(int i = 0; i < 4; i++)26 num = num*10 + d[i];27 if(vis[num]) return false;28 vis[num] = true;29 return true;30 }31 bool ok(int *d){32 for(int i = 0; i < 4; i++)33 if(d[i] != tar[i]) return false;34 return true;35 }36 int bfs(){37 while(!q.empty()) q.pop();38 node a,b;39 memcpy(a.key,temp,sizeof(temp));40 memset(vis,false,sizeof(vis));41 a.step = 0;42 q.push(a);43 transformIt(a.key);44 while(!q.empty()){45 b = q.front();46 q.pop();47 if(ok(b.key)) return b.step;48 for(int i = 0; i < 4; i++){49 a.step = b.step+1;50 memcpy(temp,b.key,sizeof(temp));51 if(temp[i] + 1 > 9) temp[i] = 1;52 else temp[i]++;53 if(transformIt(temp)){54 memcpy(a.key,temp,sizeof(temp));55 q.push(a);56 }57 memcpy(temp,b.key,sizeof(temp));58 if(temp[i]-1 == 0) temp[i] = 9;59 else temp[i]--;60 if(transformIt(temp)){61 memcpy(a.key,temp,sizeof(temp));62 q.push(a);63 }64 if(i){65 memcpy(temp,b.key,sizeof(temp));66 swap(temp[i],temp[i-1]);67 if(transformIt(temp)){68 memcpy(a.key,temp,sizeof(temp));69 q.push(a);70 }71 }72 if(i < 3){73 memcpy(temp,b.key,sizeof(temp));74 swap(temp[i],temp[i+1]);75 if(transformIt(temp)){76 memcpy(a.key,temp,sizeof(temp));77 q.push(a);78 }79 }80 }81 }82 return -1;83 }84 int main(){85 int t,i,j;86 char md[6];87 scanf("%d",&t);88 while(t--){89 scanf("%s",md);90 for(i = 0; i < 4; i++) temp[i] = md[i] - ‘0‘;91 scanf("%s",md);92 for(i = 0; i < 4; i++) tar[i] = md[i] - ‘0‘;93 printf("%d\n",bfs());94 }95 return 0;96 }
HDU 1195 Open the Lock
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