POJ - 1222 EXTENDED LIGHTS OUT高斯消元

Description

In an extended version of the game Lights Out, is a puzzle with 5 rows of 6 buttons each (the actual puzzle has 5 rows of 5 buttons each). Each button has a light. When a button is pressed, that button and each of its (up to four) neighbors above, below, right and left, has the state of its light reversed. (If on, the light is turned off; if off, the light is turned on.) Buttons in the corners change the state of 3 buttons; buttons on an edge change the state of 4 buttons and other buttons change the state of 5. For example, if the buttons marked X on the left below were to be pressed,the display would change to the image on the right. 

The aim of the game is, starting from any initial set of lights on in the display, to press buttons to get the display to a state where all lights are off. When adjacent buttons are pressed, the action of one button can undo the effect of another. For instance, in the display below, pressing buttons marked X in the left display results in the right display.Note that the buttons in row 2 column 3 and row 2 column 5 both change the state of the button in row 2 column 4,so that, in the end, its state is unchanged. 

Note: 
1. It does not matter what order the buttons are pressed. 
2. If a button is pressed a second time, it exactly cancels the effect of the first press, so no button ever need be pressed more than once. 
3. As illustrated in the second diagram, all the lights in the first row may be turned off, by pressing the corresponding buttons in the second row. By repeating this process in each row, all the lights in the first 
four rows may be turned out. Similarly, by pressing buttons in columns 2, 3 ?, all lights in the first 5 columns may be turned off. 
Write a program to solve the puzzle.

Input

The first line of the input is a positive integer n which is the number of puzzles that follow. Each puzzle will be five lines, each of which has six 0 or 1 separated by one or more spaces. A 0 indicates that the light is off, while a 1 indicates that the light is on initially.

Output

For each puzzle, the output consists of a line with the string: "PUZZLE #m", where m is the index of the puzzle in the input file. Following that line, is a puzzle-like display (in the same format as the input) . In this case, 1's indicate buttons that must be pressed to solve the puzzle, while 0 indicate buttons, which are not pressed. There should be exactly one space between each 0 or 1 in the output puzzle-like display.

Sample Input

2
0 1 1 0 1 0
1 0 0 1 1 1
0 0 1 0 0 1
1 0 0 1 0 1
0 1 1 1 0 0
0 0 1 0 1 0
1 0 1 0 1 1
0 0 1 0 1 1
1 0 1 1 0 0
0 1 0 1 0 0

Sample Output

PUZZLE #1
1 0 1 0 0 1
1 1 0 1 0 1
0 0 1 0 1 1
1 0 0 1 0 0
0 1 0 0 0 0
PUZZLE #2
1 0 0 1 1 1
1 1 0 0 0 0
0 0 0 1 0 0
1 1 0 1 0 1
1 0 1 1 0 1

Source

Greater New York 2002

这题可以利用枚举法做，本人比较懒就不打了，我写的是高斯消元的方法，就是线性代数中一个解方程组的方法。

就是说我们可以把本题看成是，灯一开始是全灭的，经过数次按键之后达到我们想要的状态，对于每一个灯每一个按扭有可能使其转换状态，也可能不能，那么能转换的记为1，不能的记为0，相当于ax+by+cz.....=l,l表示灯的状态（记得用亦或运算），这样对每一个灯我们就可以列出一个1*31的方程，总共有三十个灯就可以得到一个30*31的方程组，然后就是利用高斯消元解方程组了。解出来的那些系数就是结果。

下面是我的代码，因为我觉得本题每一个按键按与不按都是确定的，所以把高斯校园标准的模板中的一些东西删去了。但是是可以AC的。至于 完整的模板你们可以上网找，作为水牛的我就不提供了。

#include <stdio.h>
#include<algorithm>
#include <string.h>
using namespace std;

int a[40][40];
int ans[40];

void init()
{
	memset(a,0,sizeof(a));
	int i,j,k;
	for(i=0;i<5;i++)
		for(j=0;j<6;j++)
		{
			k=i*6+j;
			a[k][k]=1;
			if(i>0)
				a[k-6][k]=1;
			if(i<4)
				a[k+6][k]=1;
			if(j>0)
				a[k-1][k]=1;
			if(j<5)
				a[k+1][k]=1;
		}
}

void Gauss()
{
	int i,j,k,l;
	for(i=0;i<30;i++)
	{
		for(j=i;j<30;j++)
		{
			if(a[j][i]==1)
				break;
		}
		if(j!=i)
			for(k=0;k<=30;k++)
				swap(a[i][k],a[j][k]);
		for(k=j+1;k<30;k++)
			if(a[k][i])
				for(l=0;l<=30;l++)
					a[k][l]^=a[i][l];
	}
	for(i=29;i>=0;i--)
	{
		ans[i]=a[i][30];
		for(j=i+1;j<30;j++)
		{
			ans[i]^=(a[i][j]&&ans[j]);
		}
	}
}

int main()
{
	int t;
	scanf("%d",&t);
	int icase;
	for(icase=1;icase<=t;icase++)
	{
		init();
		int i,j,k;
		for(i=0;i<30;i++)
			scanf("%d",&a[i][30]);
		Gauss();
        printf("PUZZLE #%d",icase); 
		for(i=0;i<30;i++)
		{
			if(i%6==0)
				printf("\n%d",ans[i]);
			else
				printf(" %d",ans[i]);
		}
       printf("\n");
	}
	return 0;
}