学步园

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.

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.

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.

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

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

Greater New York 2002

前两天去旅游了，今天继续学习，收收心~~~
这题和1681一样，都是利用高消来解异或方程组。构造系数矩阵mat[i][j],如果按下i对j有影响，那么就为1，否则是0.

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;

int free_x[35];
int x[35];
int mat[35][35];

int Gauss(int equ, int var)
{
	int free_num,num=0;
	int free_index;
	int i, j, k;
	int max_r;
	int col = 0;
	for(i = 0;i < var + 1; i++)
    	x[i] = 0;
	for ( k = 0; k < equ && col < var; k++, col++)
	{
		max_r = k;
		for ( i = k+1; i < equ ; i++)
		{
			if(abs(mat[i][col]) > abs(mat[max_r][col]))
			  max_r = i;
		}
		if(max_r != k)
		{
			for(i = k; i< var + 1; i++)
			  swap(mat[k][i], mat[max_r][i]);
		}
		if(mat[k][col] == 0)
		{
			k--;
			free_x[num++]=col;
			continue;
		}
		for(i = k+1; i < equ; i++)//消元
		{
			if(mat[i][col])
			  for(j = col; j < var + 1; j++)
			     mat[i][j] ^= mat[k][j];
		}
	}
	for(i = k; i < equ; i++)
	  if(mat[i][var])
	    return -1;
    if(k < var)
      return var - k;
	for(i = var - 1; i >= 0; i--)
	{
		int temp = mat[i][var];
		for(j = i + 1;j < var; j++)
		  temp ^= x[j] * mat[i][j];
        x[i] = temp;
	}
	return 0;
}

int main()
{
	int t, p=1;
	scanf("%d", &t);
	while(t--)
	{
		memset(mat, 0, sizeof(mat));
		for(int i = 0; i < 5; i++)
		  for(int j = 0; j < 6; j++)
		  {
  			 int d = i * 6 + j;
  			 mat[d][d] = 1;
  			 if(i > 0)
  			  mat[(i - 1) * 6 + j][d] = 1;
		     if(i < 4)
		      mat[(i + 1) * 6 + j][d] = 1;
	         if(j > 0)
	          mat[i * 6 + j - 1][d] = 1;
             if(j < 5)
              mat[i * 6 + j + 1][d] = 1;
  		  }
		for(int i = 0; i < 30; i++)
  			scanf("%d", &mat[i][30]);
        int res = Gauss(30, 30);
        printf("PUZZLE #%d\n", p++);
        for(int i = 0; i< 5; i++)
        {
        	for(int j = 0 ;j < 5; j++)
        	  printf("%d ", x[i * 6 + j]);
      	    printf("%d\n", x[i * 6 + 5]);
        }        
	}
	return 0;
}