http://poj.org/problem?id=2965

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

这道题本以为用位运算和BFS就可以搞定的。。。但是先是WA几次，然后就是RTE,,,那个杯具啊，，，最后看到一位大牛的总结，结论是如果按题目的规则，进行操作，在把+变成-的情况下，不改变其他原来的状态，只需把当前为+的位置handle7次该+号所在行列的其他元素handle6次，其他元素handle4次即可，根据这一结论，给出高效算法，就是把+号所在的行和列的位置自增一次，最后遍历该数组，找出为奇数的位置，则为要操作的位置，而所有为奇数之和则为需要操作的次数。

代码：

#include<iostream>
using namespace std;
int a[4][4];
int main()
{   for(int i=0;i<4;i++)
      for(int j=0;j<4;j++)
       {       char s;
              cin>>s;
              if(s=='+')
               { for(int p=0;p<4;p++)
                    { a[i][p]++;
                      a[p][j]++;
                      }
                      a[i][j]--;
              }
              }
       int sum=0;
       for(int i=0;i<4;i++)
         for(int j=0;j<4;j++)
          if(a[i][j]%2) sum++;
           cout<<sum<<endl;
          for(int i=0;i<4;i++)
           for(int j=0;j<4;j++)
            if(a[i][j]%2)
            cout<<i+1<<" "<<j+1<<endl;
            return 0;
    }