The Pilots Brothers' refrigerator 枚举 Poj 2965

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本文介绍了一个基于矩阵操作的游戏谜题解决方案。玩家需要通过改变冰箱门上16个把手的状态（开或关）来打开冰箱。文章提供了一种算法，通过最少步骤将所有把手调整到开启状态，包括代码实现。

The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24669		Accepted: 9510		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

Source

题意：　有一个冰箱，门上有４＊４，16个把手，＇+＇代表关闭，＇-＇代表打开．现在给出这个４＊４的把手矩阵，表示把手的状态，问至少需要几步可以把冰箱打开（把手状态全部变为＇－＇），求step?

题解思路：

奇变偶不变，把手状态变化偶数倍和不变化是一样的，用一个表变化次数的矩阵mp标记变化次数，最后根据是否变化奇数次来判断输出．

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
int mp[4][4];

int main()
{
    int step=0;
    char a;
    memset(mp, 0, sizeof(mp));
    for(int i = 0; i < 4; i++)
    {
        for(int j = 0; j < 4; j++)
        {
            scanf("%c", &a);
            if(a == '+')
            {
                mp[i][j]++;
                for(int k = 0; k < 4; k++)
                {
                    mp[i][k]++;
                    mp[k][j]++;
                }
            }
        }
        getchar();       //*****注意点*****
    }
    for(int i = 0; i < 4; i++)
    {
        for(int j = 0; j < 4; j++)
        {
            if(mp[i][j]%2 == 1)
                step++;
        }
    }
    printf("%d\n", step);
    for(int i = 0; i < 4; i++)
    {
        for(int j = 0; j < 4; j++)
        {
            if(mp[i][j]%2 == 1)
                printf("%d %d\n", i+1, j+1);
        }
    }
    return 0;
}