POJ 1222 EXTENDED LIGHTS OUT (高斯消元)

EXTENDED LIGHTS OUT

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8364		Accepted: 5406

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




题意：给你一个5*6的图，然后每一块上有一个灯，当你点击一个点的开关，它的周围四个方向的开关也会变化，可以
点击多个开关，问应该点击哪几个

思路：高斯消元的模板，其实就是线性代数中普通矩阵转化为阶梯型矩阵的过程，相当于一个方程组，一般是5个变元
，5个方程，然后根据它的增广矩阵来求此方程组是否有解，困难的地方在于方程组的建立，即初始矩阵的建立，这个
题目就想当于x(1,1)*a1+x(1,2)*a2+....+x(5,6)*a30=0这样一个方程组，其中有30个方程式，建立增广矩阵求解，其
实我对高斯消元理解的也不是很透彻，还需要多多努力。。。

ac代码： 
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int a[MAXN][MAXN];
int equ,var;
int x[MAXN];
int free_x[MAXN];
int free_num;
void debug()
{
	int i,j;
	printf("debug\n");
	for(i=0;i<equ;i++)
	{
		for(j=0;j<var;j++)
		printf("%d ",a[i][j]);
		printf("\n");
	}
}
int Gauss()
{
	int i,j,col,k,max_r;
	free_num=0;
	for(k=0,col=0;k<equ&&col<var;k++,col++)
	{
		max_r=k;
		for(i=k+1;i<equ;i++)
		{
			if(abs(a[i][col])>abs(a[max_r][col]))
			max_r=i;
		}
		if(a[max_r][col]==0)
		{
			k--;
			free_x[free_num++]=col;
			continue;
		}
		if(max_r!=k)
		{
			for(j=col;j<var+1;j++)
			swap(a[k][j],a[max_r][j]);
		}
		for(i=k+1;i<equ;i++)
		{
			if(a[i][col])
			{
				int lcm=a[k][col]/gcd(a[k][col],a[i][col])*a[i][col];
                int ta=lcm/a[i][col],tb=lcm/a[k][col];
                if(a[i][col]*a[k][col]<0)
				tb=-tb;
                for(int j=col;j<var+1;j++)
                a[i][j]=((a[i][j]*ta)%2-(a[k][j]*tb)%2+2)%2;
			}
		}
	}
	for(i=k;i<equ;i++)
	if(a[i][col])
	return -1;
	if(k<var)
	return var-k;
	for(i=var-1;i>=0;i--)
	{
		int temp=a[i][var]%2;
		for(j=i+1;j<var;j++)
		if(a[i][j])
		temp=(temp-a[i][j]*x[j]%2+2)%2;
		x[i]=(temp/a[i][i]%2);
	}
	return 0;
}
void init()
{
	int i,j;
	mem(a);mem(x);mem(free_x);
	equ=30;var=30;
	for(i=0;i<5;i++)
	{
		for(j=0;j<6;j++)
		{
			if(i) 
			a[i*6+j][(i-1)*6+j]=1;  
            if(i!=4) 
		    a[i*6+j][(i+1)*6+j]=1;  
            if(j) 
		    a[i*6+j][i*6+j-1]=1;  
            if(j!=5) 
		    a[i*6+j][i*6+j+1]=1;  
            a[i*6+j][i*6+j]=1;
		}
	}
	for(i=0;i<30;i++)
	scanf("%d",&a[i][30]);
}
int main()
{
	int t,i,j;
	int cas=0;
	scanf("%d",&t);
	while(t--)
	{
		init();
		Gauss();
		printf("PUZZLE #%d\n",++cas);
		for(i=0;i<5;i++)
		{
			printf("%d",x[i*6]);
			for(j=1;j<6;j++)
			printf(" %d",x[i*6+j]);
			printf("\n");
		}
	}
	return 0;
}