（枚举，dfs）The Pilots Brothers' refrigerator poj 2965

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本文介绍了一款游戏中的挑战任务——通过最少的操作次数将冰箱的所有把手从关闭状态切换到打开状态。文章详细解释了游戏机制，并提供了一段C++代码实现，该算法能够找出解决这一问题的最优解。

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

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[15][15];
int a[15][15];
int alt[15][15];
int num;
int ans[15][15];
int judge()
{
    for(int i=1; i<=4; ++i)
        for(int j=1; j<=4; ++j)
        {
            if(!a[i][j])
                return 0;
        }
    return 1;
}
void fil(int x,int y)
{
    alt[x][y]=!alt[x][y];
    for(int i=1; i<=4; ++i)
        a[x][i]=!a[x][i];
    for(int i=1; i<=4; ++i)
        a[i][y]=!a[i][y];
    a[x][y]=!a[x][y];
}
void copyans()
{
    for(int i=1; i<=4; ++i)
        for(int j=1; j<=4; ++j)
            ans[i][j]=alt[i][j];
}
int dfs(int x,int y,int cnt)
{
    if(judge())
    {
        num=min(num,cnt);
        copyans();
        return 0;
    }
    if(x>4)
        return 0;
    int ny=(y+1)%5;
    int nx=x+(y+1)/5;
    if(ny==0)
        ny=1;
    dfs(nx,ny,cnt);

    fil(x,y);
    dfs(nx,ny,cnt+1);
    fil(x,y);
    return 0;
}
void put()
{
    for(int i=1; i<=4; ++i)
        for(int j=1; j<=4; ++j)
            if(ans[i][j])
                cout<<i<<" "<<j<<endl;
}
int main()
{
    for(int i=1; i<=4; ++i)
        scanf("%s",s[i]+1);

    for(int i=1; i<=4; ++i)
        for(int j=1; j<=4; ++j)
        {
            if(s[i][j]=='+')
                a[i][j]=0;
            else
                a[i][j]=1;
        }
    memset(alt,0,sizeof(alt));
    num=999999;
    dfs(1,1,0);
    cout<<num<<endl;
    put();
}