标签：规律 int 题解 行和列 handles POJ2965 open refrigerator 翻转

POJ2965(找规律)题解

题目

The Pilots Brothers’ refrigerator
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 35766 Accepted: 13828

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.

题意

这题和POJ1753差不多，只是每一次改变影响的范围变成了行和列的所有点，最终目的是全部变为“-”。

分析

一开始，我直接套用的我之前写的 POJ1753(枚举)的代码，更改了翻转函数，加了个string用于存储翻转过程中的i和j。结果超时了，但最多只有65535次运算，不知道为毛会超时。
后来转换思路，发现了规律，想要将一个点s的“+”变成“-”，只要将s所在的行和列的所有点都翻转一遍就能实现。因为整个过程中只有s点翻转了奇数次，s所在的行和列的其他点都翻转了偶数次，等于没变化。所以我们只需要遍历矩阵，每次遇到“+”就翻转改点所在的行和列的所有点一次。统计所有点翻转的次数，如果是奇数则要翻转，偶数则不用。输出所有要翻转的点。

代码

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
	int g[7][7] = {0};
	for (int i = 1; i <= 4; i ++ )
	{
		for (int j = 1; j <= 4; j ++ )
		{
			char c;
			cin >> c;
			if (c == '+')
			{
				for (int k = 1; k <= 4; k ++ )
				{
					g[i][k] ++;
					g[k][j] ++;
				}
				g[i][j] --;
			}
		}
	}
	int res = 0;
	for (int i = 1; i <= 4; i ++ )
	{
		for (int j = 1; j <= 4; j ++ )
		{
			if (g[i][j] & 1) res ++ ;
		}
	}
	cout << res << endl;
	for (int i = 1; i <= 4; i ++ )
	{
		for (int j = 1; j <= 4; j ++ )
		{
			if (g[i][j] & 1) cout << i << " " << j << endl;
		}
	}
	return 0;
}

标签：规律,int,题解,行和列,handles,POJ2965,open,refrigerator,翻转
来源： https://blog.csdn.net/weixin_45609500/article/details/119452604

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