poj 2965--The Pilots Brothers' refrigerator
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
题意跟上篇poj1753类似，不同的是此题，翻转一个的同时，会改变这个所在的行和列上所有元素的状态，另外该题输出的是最后的改变路径，只要将1753的翻转函数改一下，并在搜索时加上两个数组记录路径即可，大致思路相同。
View Code
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <queue>
5 #include <stack>
6 using namespace std;
7
8 int map[10][10],n,flag;
9 int dx[20],dy[20];
10 void filp(int row,int col)
11 {
12 int j;
13 for(j=0;j<4;j++)
14 {
15 map[row][j]=!map[row][j];
16 map[j][col]=!map[j][col];
17 }
18 map[row][col]=!map[row][col];
19 }
20 int judge()
21 {
22 int i,j;
23 for(i=0;i<4;i++)
24 for(j=0;j<4;j++)
25 if(map[i][j]==0)
26 return 0;
27 return 1;
28 }
29 void dfs(int row,int col,int cnt)
30 {
31 if(cnt==n)
32 {
33 flag=judge();
34 return ;
35 }
36 if(flag||row>=4)
37 return;
38 if(row<4)
39 {
40 if(col<3)
41 {
42 filp(row,col);
43 dx[cnt]=row,dy[cnt]=col;
44 dfs(row,col+1,cnt+1);
45 if(flag) return;
46 filp(row,col);
47 dfs(row,col+1,cnt);
48 }
49 else
50 {
51 filp(row,col);
52 dx[cnt]=row,dy[cnt]=col;
53 dfs(row+1,0,cnt+1);
54 if(flag) return;
55 filp(row,col);
56 dfs(row+1,0,cnt);
57 }
58 }
59 else
60 return;
61 }
62 int main()
63 {
64 char s[5][5];
65 int i,j;
66 flag=0;
67 for(i=0;i<4;i++)
68 {
69 scanf("%s",s[i]);
70 for(j=0;j<4;j++)
71 {
72 if(s[i][j]=='-')
73 map[i][j]=1;
74 else
75 map[i][j]=0;
76 }
77 }
78 for(n=0;n<=16;n++)
79 {
80 dfs(0,0,0);
81 if(flag)
82 break;
83 }
84 printf("%d\n",n);
85 for(i=0;i<n;i++)
86 printf("%d %d\n",dx[i]+1,dy[i]+1);
87 return 0;
88 }
