本文主要是介绍poj 2965 The Pilots Brothers' refrigerator——DFS（分类是枚举），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 17929 | Accepted: 6794 | 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
Source
这个题和1753那个题很像，无非就是1753翻的时候是周围的都翻，而这个是翻得时候它所在的行和列都跟着翻（有个坑）...
#include <iostream>
#include <cstdlib>
#include <cstring>using namespace std;char mp1[4][4];
int mp2[4][4];struct node
{int x;int y;
} ls[100010];int pd()
{int i,j,js=0;for(i=0; i<4; i++){for(j=0; j<4; j++){if(mp2[i][j]==1){js++;}}}if(js==16)return 1;return 0;
}void fan(int x,int y)
{int i;for(i=0; i<4; i++){if(mp2[x][i]==1)mp2[x][i]=0;elsemp2[x][i]=1;}for(i=0; i<4; i++){if(mp2[i][y]==1)mp2[i][y]=0;elsemp2[i][y]=1;}if(mp2[x][y]==1)//这里是个坑，上面行翻得时候mp2[x][y]，已经翻过mp2[x][y]=0;//了，列翻的时候又给翻回去了,所以这里要给翻回来elsemp2[x][y]=1;
}int dfs(int x,int y,int t)
{int bx,by,i;if(pd()){cout<<t<<endl;for(i=0; i<t; i++){cout<<ls[i].x<<" "<<ls[i].y<<endl;}return 0;}if(x>=4||y>=4){return 0;}bx=(x+1)%4;by=y+(x+1)/4;dfs(bx,by,t);fan(x,y);ls[t].x=x+1;ls[t].y=y+1;dfs(bx,by,t+1);fan(x,y);
}int main()
{int i,j;for(i=0; i<4; i++){cin>>mp1[i];}for(i=0; i<4; i++){for(j=0; j<4; j++){if(mp1[i][j]=='+'){mp2[i][j]=0;}else{mp2[i][j]=1;}}}dfs(0,0,0);return 0;
}
如果不明白，去看看1753那个题的题解吧....http://blog.csdn.net/codeblf2/article/details/31040891
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