poj2965 The Pilots Brothers' refrigerator

最新推荐文章于 2024-05-05 08:00:34 发布

kimnoic

最新推荐文章于 2024-05-05 08:00:34 发布

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本文介绍了一个基于位运算的编程问题——如何通过最少的操作打开一个由16个手柄组成的冰箱门。每个操作会改变指定行和列的所有手柄状态。文章提供了一段C++代码示例，展示了如何通过暴力搜索找到最优解。

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位运算暴搜

The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20188		Accepted: 7764		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

题意为一个4*4 的一个棋盘，每次变动一个子会使这个子所在的行和列都改变，为最少变动几个子可以使整个棋盘变成一种棋子

同样只有2^16种情况，每种情况对应只有16种状态。先把这16种翻转状态求出来，再从头暴搜，找出最小记录。

#include <iostream>
using namespace std;
int po[16] = {1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768};
int a[4][4]={};
int main()
{
	for(int i=0;i<4;i++)
	{
		for(int j=0;j<4;j++)
		{
			for(int k=4*i;k<4*i+4;k++) a[i][j]+=po[k];
			for(int k=j;k<16;k+=4) a[i][j]+=po[k];
			a[i][j]-=po[4*i+j];
		}
	}
	char tmp;
	int t=0;
	for(int i=0;i<16;i++)
	{
		cin>>tmp;
		if(tmp=='+') t+=po[i];
	}
	int res=17;
	int e;
	for(int k=0;k<65536;k++)
	{
		int v=t;
		int cnt=0;
		for(int i=0;i<4;i++)
		{
			for(int j=0;j<4;j++)
			{
				if(k&po[4*i+j])
				{
					cnt++;
					v^=a[i][j];
				}
			}
		}
		//cout<<cnt<<endl;
		if(v==0)
		{
			if(res>cnt)
			{
				res=cnt;
				e=k;
			}
		}
	}
	cout<<res<<endl;
	for(int i=0;i<4;i++)
	{
		for(int j=0;j<4;j++)
		{
			if(e&po[4*i+j])
			{
				cout<<i+1<<" "<<j+1<<endl;
			}
		}
	}
	return 0;
}