HDU 1195 Open the Lock

时间：2014-08-29 19:56:58 阅读：158 评论：0 收藏：0 [点我收藏+]

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Open the Lock

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 1195
64-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.

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

Sample Output

2
4

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 long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 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 }

View Code

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 long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 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 }

View Code

HDU 1195 Open the Lock

标签：style blog http color os java io ar for

原文地址：http://www.cnblogs.com/crackpotisback/p/3945570.html

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