The Pilots Brothers' refrigerator

最新推荐文章于 2024-09-28 22:31:50 发布

原创最新推荐文章于 2024-09-28 22:31:50 发布 · 216 阅读

0 ·

CC 4.0 BY-SA版权

poj 专栏收录该内容

25 篇文章

订阅专栏

本文介绍了一个基于矩阵操作的游戏谜题——打开带有16个把手的冰箱门。通过切换特定位置的把手状态，同时影响同一行和列的其他把手，目标是最小化开关次数以打开所有把手。

The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24421		Accepted: 9416		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

Northeastern Europe 2004, Western Subregion

#include<iostream>
#include<cstring>
using namespace std;
char map[4][4];
bool flag[4][4];
int ans = 0;
int main()
{
	int si[16],sj[16];
	memset(flag,0,sizeof(flag));
	for(int i = 0;i < 4;i++)
	{
		cin >> map[i];
	}
	for(int i = 0;i < 4;i++)
	{
		for(int j = 0;j < 4;j++)
		{
			char e = map[i][j];
			if(e == '+')
			{
				flag[i][j] = !flag[i][j];
				for(int k = 0;k < 4;k++)
				{
					flag[i][k] = !flag[i][k];
					flag[k][j] = !flag[k][j];
				}
			}
		}
	}
	for(int i = 0;i <  4;i++)
	{
		for(int j = 0;j < 4;j++)
		{
			if(flag[i][j] == true)
			{
				si[ans] = i+1;
				sj[ans] = j+1;
				ans++;
			}
		}
	}
	cout<<ans<<endl;
	for(int i = 0;i < ans;i++)
	{
		cout<<si[i]<<" "<<sj[i]<<endl;
	}
	return 0;
}