poj 1222 EXTENDED LIGHTS OUT （高斯消元法）

最新推荐文章于 2020-12-31 13:40:43 发布

hncu__lz

最新推荐文章于 2020-12-31 13:40:43 发布

阅读量434

点赞数

CC 4.0 BY-SA版权

分类专栏：高斯消元法

本文链接：https://blog.youkuaiyun.com/hndu__lz/article/details/60599548

高斯消元法专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种解决LightsOut游戏的算法，通过将其转化为0、1矩阵方程组，并使用高斯消元法求解未知数，从而找到需要按下的按钮以熄灭所有灯。

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

题意：有一个5*6的矩阵，每个位置都表示按钮和灯，1表示亮，0表示灭。每当按下一个位置的按钮，它和它周围灯的状态全部翻转，问在这样的一个方阵中按下哪些按钮可以把整个方阵都变成灭的，这时1表示按了，0表示没按。

思路：转自：http://blog.youkuaiyun.com/shiren_Bod/article/details/5766907

这个游戏有一些技巧：
1、按按钮的顺序可以随便。
2、任何一个按钮都最多需要按下1次。因为按下第二次刚好抵消第一次，等于没有按。

这个问题可以转化成数学问题。
一个灯的布局可以看成一个0、1矩阵。以3x3为例：
0 1 0
1 1 0
0 1 1
表示一个布局。其中0表示灯灭，1表示灯亮。
每次按下按钮（POJ1222）或者叫一个宿舍关灯（0998），可以看成在原矩阵上加（模2加，就是按位异或）上一个如下的矩阵：
0 1 0
1 1 1
0 1 0
上述矩阵中的1表示按下第2行第2列的按钮时，作用的范围。如果按左上角的按钮，就是：
1 1 0
1 0 0
0 0 0

我们记L为待求解的原始布局矩阵。A(i,j)表示按下第i行第j列的按钮时的作用范围矩阵。在上述例子中，
L=
0 1 0
1 1 0
0 1 1

A(1,1)=
1 1 0
1 0 0
0 0 0

A(2,2)=
0 1 0
1 1 1
0 1 0

假设x(i,j)表示：想要使得L回到全灭状态，第i行第j列的按钮是否需要按下。0表示不按，1表示按下。那么，这个游戏就转化为如下方程的求解：
L + x(1,1)*A(1,1) + x(1,2)*A(1,2) + x(1,3)*A(1,3) + x(2,1)*A(2,1) + ... + x(3,3)*A(3,3) = 0

其中x(i,j)是未知数。方程右边的0表示零矩阵，表示全灭的状态。直观的理解就是：原来的L状态，经过了若干个A(i,j)的变换，最终变成0:全灭状态。
由于是0、1矩阵，上述方程也可以写成：
x(1,1)*A(1,1) + x(1,2)*A(1,2) + x(1,3)*A(1,3) + x(2,1)*A(2,1) + ... + x(3,3)*A(3,3) = L

这是一个矩阵方程。两个矩阵相等，充要条件是矩阵中每个元素都相等。将上述方程展开，便转化成了一个9元1次方程组：

简单地记做：AA * XX = LL

这个方程有唯一解：
x(1,1) x(1,2) x(1,3)
x(2,1) x(2,2) x(2,3)
x(3,1) x(3,2) x(3,3)
=
1 1 1
0 0 0
0 0 1

也就是说，按下第一行的3个按钮，和右下角的按钮，就

能使L状态变成全灭状态。
对于固定行列的阵列来说，AA矩阵也是确定的。是否存在解，解是否唯一，只与AA矩阵有关。对于唯一解的情形，只要将LL乘以AA的逆矩阵即可。具体求AA的逆矩阵的方法，可以用高斯消元法。

由于是0、1矩阵，上述方程也可以写成：

将1式两边同时加上一个L矩阵就可以变成
x(1,1)*A(1,1) + x(1,2)*A(1,2) + x(1,3)*A(1,3) + x(2,1)*A(2,1) + ... + x(3,3)*A(3,3) = L

A（1，1）把矩阵转化为一个列向量，L也转化为一个列向量，

将sigma xi*Ai=Li 对应位置的值相等就可以建立方程组了

X1*A(1,1)1+X2*A(1,2)1+X3*A(1,3)1+…………X30*A(30,30)1=L1; mod 2

X1*A(1,1)2+X2*A(1,2)2+X3*A(1,3)2+…………X30*A(30,30)2=L2; mod 2

X1*A(1,1)3+X2*A(1,2)3+X3*A(1,3)3+…………X30*A(30,30)3=L3 mod 2

…….

X1*A(1,1)30+X2*A(1,2)30+X3*A(1,3)30+…………X30*A(30,30)30=L30; mod 2

其中A(i,j)k 表示列向量A中第K个元素

这里的*表示点乘，Xi取(1,0) +表示模2加法，所以在高斯消元的时候可以用^异或运算

1个开关最多控制5个灯，在构造的矩阵中，a[i][j]=1表示第i个开关可以影响到j号灯

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[30][35];
int equ=30,var=30;
int free_x[30],x[30];
void init()
{
    memset(a,0,sizeof(a));
    for(int i=0;i<5;i++)
        for(int j=0;j<6;j++)
        {
            int t=i*6+j;
            a[t][t]=1;
            if(i>0) a[(i-1)*6+j][t]=1;
            if(i<4) a[(i+1)*6+j][t]=1;
            if(j>0) a[i*6+j-1][t]=1;
            if(j<5) a[i*6+j+1][t]=1;
        }
}
int gcd(int a, int b)
{
    int t;
    while (b != 0)
    {
        t = b;
        b = a % b;
        a = t;
    }
    return a;
}
inline int lcm(int a, int b)
{
    return a * b / gcd(a, b);
}
int Gauss()
{
    int i,j,k;
    int max_r;
    int col;
    int ta,tb;
    int LCM;
    int temp;
    int free_x_num;
    int free_index;
    for(k=0,col=0;k<30,col<30;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(max_r!=k)
            for(j=k;j<var+1;j++) swap(a[k][j],a[max_r][j]);
        if(a[k][col]==0)
        {
            k--;continue;
        }
        for(i=k+1;i<equ;i++)
        {
            if(a[i][col]!=0)
            {
                /*LCM=lcm(abs(a[i][col]),abs(a[k][col]));
                ta=LCM/abs(a[i][col]),tb=LCM/abs(a[k][col]);
                if(a[i][col]*a[k][col]<0) tb=-tb;
                for(j=col;j<var+1;j++)
                    a[i][j]=a[i][j]*ta-a[k][j]*tb;
                */
                for(j=col;j<var+1;j++)
                    a[i][j]^=a[k][j];
            }
        }
    }
    /*for(i=k;i<equ;i++)
        if(a[i][col]!=0) return -1;
    printf("%d %d\n",k,var);
    if(k<var)
    {
        for(i=k-1;i>=0;i--)
        {
            free_x_num=0;
            for(j=0;j<var;j++)
                if(a[i][j]!=0&&!free_x[j])
                {
                    free_x_num++;
                    free_index=j;
                }
            if(free_x_num>1) continue;
            temp=a[i][var];
            for(j=0;j<var;j++)
                if(a[i][j]!=0&&j!=free_index)
                {
                    temp-=x[j]*a[i][j];
                }
            x[free_index]=temp/a[i][free_index];
            free_x[free_index]=1;
        }
        return var-k;
    }
    */
    for(i=var-1;i>=0;i--)
    {
        /*temp=a[i][var];
        for(j=i+1;j<var;j++)
            if(a[i][j]!=0) temp-=a[i][j]*x[j];*/
        x[i]=a[i][30];
        for(j=i+1;j<var;j++)
            x[i]^=(a[i][j]&&x[j]);
    }
    return 0;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int t=1;t<=T;t++)
    {
        memset(free_x,0,sizeof(free_x));
        memset(x,0,sizeof(x));
        init();
        for(int i=0;i<30;i++)
            scanf("%d",&a[i][30]);
        int free_num=Gauss();
        printf("PUZZLE #%d\n",t);
        for(int i=0;i<5;i++)
        {
            for(int j=0;j<6;j++)
            {
                if(j==5){ printf("%d",x[i*6+j]);break;}
                printf("%d ",x[i*6+j]);
            }
            printf("\n");
        }
    }
}