POJ The Pilots Brothers' refrigerator 2965-优快云博客

The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23898		Accepted: 9208		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 iand 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

题意：翻动棋子，把棋子都变成－，翻动某个棋子，他所在的行列都翻动，求最小的步数和翻动先后坐标位置

其实这个题用枚举＋dfs能做，可惜我不会

在网上看的，如果想把一个＋变成－，就翻动他所在这一行和这一列的所有棋子，可以试一下

我们就遍历，找到＋，就把这一行和这一列都加１，表示翻动了１次，一个棋子翻动偶数次等于没翻

我们就把翻动次数取余２，这样就去掉重复翻动的次数

可以去网上找那个大神讲的

</pre><pre name="code" class="cpp">#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
struct node
{
    int x,y;
}q[20];
int main()
{
    char mp[5][5];
    int mpp[5][5];
    memset(mpp,0,sizeof(mpp));
    int i,j,x,y;
    for(i=0;i<4;i++)
    {
        for(j=0;j<4;j++)
        {
            cin>>mp[i][j];
            if(mp[i][j]=='+')
            {
                mpp[i][j]++;
                for(x=0;x<4;x++)
                {
                    mpp[i][x]++;
                    mpp[x][j]++;
                }
            }
        }
    }
    int top=0;
    for(i=0;i<4;i++)
    {
        for(j=0;j<4;j++)
        {
            mpp[i][j]%=2;
            if(mpp[i][j]==1)
            {
                q[top].x=i+1;
                q[top].y=j+1;
                top++;
            }
        }
    }
    printf("%d\n",top);
    for(i=0;i<top;i++)
    {
        printf("%d %d\n",q[i].x,q[i].y);
    }

    return 0;
}