POJ_2965 The Pilots Brothers' refrigerator 搜索

题目链接：http://poj.org/problem?id=2965

简单搜索，一个一个往下找判断即可。同样最多16次即可，再反转同样的无意义。

代码：

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <iostream>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <stack>
 8 #include <vector>
 9 using namespace std;
10 int n = 4, maze[5][5];
11 int curstp = 0, minstp = 17;
12 int fir[17], sec[17], bfir[18], bsec[18];
13 bool flag = false;
14 
15 bool check(){
16     for(int i = 0; i < 4; i++){
17         for(int j = 0; j < 4; j++){
18             if(maze[i][j] == 0){
19                 return false;
20             }
21         }
22     }
23     return true;
24 }
25 
26 void change(int ii, int jj){
27     maze[ii][jj] = !maze[ii][jj];
28     for(int i = 0; i < 4; i++){
29         if(i != jj)
30             maze[ii][i] = !maze[ii][i];
31         if(i != ii)
32             maze[i][jj] = !maze[i][jj];
33     }
34 }
35 
36 int dfs(int t){
37     int ii = t / 4, jj = t - ii * 4;
38     if(t >= 16){
39         if(check() && curstp < minstp){
40             minstp = curstp;
41             for(int i = 0; i < 16; i++){
42                 bfir[i] = fir[i]; bsec[i] = sec[i];
43             }
44             flag = true;
45         } 
46         return 0;
47     }
48     //change 
49     if(curstp + 1 < minstp){
50         curstp++; change(ii, jj); 
51         fir[t] = ii, sec[t] = jj;
52         dfs(t + 1);
53         curstp--; change(ii, jj);
54         fir[t] = -1, sec[t] = -1;
55     }
56     //no change
57     dfs(t + 1);
58 }
59 
60 int main(){
61     for(int i = 0; i < 4; i++){
62         for(int j = 0; j < 4; j++){
63             char tmch;
64             scanf(" %c", &tmch);
65             maze[i][j] = tmch == '+'? 0: 1;
66         }
67     }
68     memset(fir, -1, sizeof(fir));
69     memset(sec, -1, sizeof(sec));
70     memset(bfir, -1, sizeof(fir));
71     memset(bsec, -1, sizeof(sec));
72     dfs(0);
73     if(flag){
74         printf("%d\n", minstp);
75         for(int i = 0; i < 16; i++){
76             if(bfir[i] != -1){
77                 printf("%d %d\n", bfir[i] + 1, bsec[i] + 1);
78             }
79         }
80     }
81     else{
82         printf("");
83     }
84 }

题目：

The Pilots Brothers' refrigerator
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 26366   Accepted: 10149   Special Judge

Description

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

Sample Input

-+--
----
----
-+--

Sample Output

6
1 1
1 3
1 4
4 1
4 3
4 4

 

posted @ 2017-05-12 09:02  EricJeffrey  阅读(127)  评论(0)  编辑  收藏  举报