POJ 1222 EXTENDED LIGHTS OUT (枚举 或者 高斯消元)

EXTENDED LIGHTS OUT

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5383		Accepted: 3557

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

 

1,枚举：

题目大意就不说了，就是把棋盘上的1全变0即可

如果枚举的话，看似有2的30次方中可能，其实不是。

实际上只需要枚举第一行的状态即可，再往后，如果想要解决问题，必须根据第一行的状态推下去。

对于每个位置，如果上一行的这一列有1，必然这个按键要按下去，不然不可能达到要求的结果。

枚举代码如下，直接使用二进制枚举

 

#include<iostream>

#include<cstdio>

#include<cstring>



using namespace std;



int map[5][6],g[5][6];

int res[5][6];



void press(int i,int j){

    g[i][j]=!g[i][j];

    if(i>0)     g[i-1][j]=!g[i-1][j];

    if(j>0)     g[i][j-1]=!g[i][j-1];

    if(i<4)     g[i+1][j]=!g[i+1][j];

    if(j<5)     g[i][j+1]=!g[i][j+1];

}



bool ok(){

    for(int j=0;j<6;j++)

        if(g[4][j])

            return 0;

    return 1;

}



void solve(){

    for(int k=0;k<64;k++){

        memcpy(g,map,sizeof(map));

        memset(res,0,sizeof(res));

        for(int j=0;j<6;j++)

            if(k&(1<<j)){

                press(0,j);

                res[0][j]=1;

            }

        for(int i=1;i<5;i++)

            for(int j=0;j<6;j++)

                if(g[i-1][j]){

                    press(i,j);

                    res[i][j]=1;

                }

        if(ok()){

            for(int i=0;i<5;i++){

                for(int j=0;j<5;j++)

                    printf("%d ",res[i][j]);

                printf("%d\n",res[i][5]);

            }

            break;

        }

    }

}



int main(){



    //freopen("input.txt","r",stdin);



    int t,cases=0;

    scanf("%d",&t);

    while(t--){

        for(int i=0;i<5;i++)

            for(int j=0;j<6;j++)

                scanf("%d",&map[i][j]);

        printf("PUZZLE #%d\n",++cases);

        solve();

    }

    return 0;

}

 

2，高斯消元

不得不说这是一种逆向思想，题目要求是怎样能全灭掉，建立方程则是求解怎样能从全灭转变到当前状态，实际上，由于一个灯摁两次的效果是一样的，所以这样求出来的解就是最终解。

然后求解的时候，最好还是用异或操作。这样就避免出现了负数和特别巨大的数字。经试验，这是很有可能的。

这就是用高斯消元法求异或方程组了

异或方程组就是形如这个样子的方程组：


M[0][0]x[0]^M[0][1]x[1]^…^M[0][N-1]x[N-1]=B[0]
M[1][0]x[0]^M[1][1]x[1]^…^M[1][N-1]x[N-1]=B[1]
…
M[N-1][0]x[0]^M[N-1][1]x[1]^…^M[N-1][N-1]x[N-1]=B[N-1]


其中“^”表示异或(XOR, exclusive or)，M[i][j]表示第i个式子中x[j]的系数，是1或者0。B[i]是第i个方程右端的常数，是1或者0。


解这种方程可以套用高斯消元法，只须将原来的加减操作替换成异或操作就可以了，两个方程的左边异或之后，它们的公共项就没有了。


具体的操作方法是这样的：对于k=0..N-1，找到一个M[i][k]不为0的行i，把它与第k行交换，用第k行去异或下面所有M[i][j]不为0的行i，消去它们的第k个系数，这样就将原矩阵化成了上三角矩阵；最后一行只有一个未知数，这个未知数就已经求出来了，用它跟上面所有含有这个未知数的方程异或，就小觑了所有的着个未知数，此时倒数第二行也只有一个未知数，它就被求出来了，用这样的方法可以自下而上求出所有未知数。

 

高斯消元介绍： http://blog.csdn.net/shiren_Bod/article/details/5766907

 

#include<iostream>

#include<cstdio>

#include<cstring>



using namespace std;



int map[5][6];

int g[35][35],res[35];

int n;



int Gauss(){

    int pos;

    for(int k=0;k<n;k++){

        pos=-1;

        for(int i=k;i<n;i++)    //找到这一列中第一个非0元素的行

            if(g[i][k]!=0){

                pos=i;

                break;

            }

        for(int i=k;i<=n;i++)

            swap(g[k][i],g[pos][i]);    //交换位置，我们在学线性代数的时候也经常这么做  

        for(int i=k+1;i<n;i++){

            if(g[i][k]==0)

                continue;

            for(int j=k;j<=n;j++)   //对于下面出现的这一列中有1的行，需要把1消掉

                g[i][j]^=g[k][j];

        }

    }

    for(int i=n-1;i>=0;i--){

        for(int j=0;j<i;j++){

            if(g[j][i]==0)

                continue;

            for(int k=0;k<=n;k++)

                g[j][k]^=g[i][k];

        }

        int flag=0;

        for(int j=0;j<n;j++)

            if(g[i][j]){

                flag=1;

                break;

            }

        if(!flag)   //如果某一行的系数全是0，说明就不对了

            return 0;

        res[i]=g[i][n];

    }

    return 1;

}



int main(){



    //freopen("input.txt","r",stdin);



    int t,cases=0;

    scanf("%d",&t);

    n=30;

    while(t--){

        memset(g,0,sizeof(g));

        for(int i=0;i<5;i++)

            for(int j=0;j<6;j++){

                scanf("%d",&map[i][j]);

                if(map[i][j])

                    g[6*i+j][n]=1;

            }

        for(int i=0;i<5;i++)

            for(int j=0;j<6;j++){

                g[i*6+j][i*6+j]=1;

                if(i>0)     g[6*i+j][6*(i-1)+j]=1;

                if(j>0)     g[6*i+j][6*i+j-1]=1;

                if(i<4)     g[6*i+j][6*(i+1)+j]=1;

                if(j<5)     g[6*i+j][6*i+j+1]=1;

            }

        printf("PUZZLE #%d\n",++cases);

        Gauss();

        for(int i=0;i<5;i++)

            for(int j=0;j<6;j++)

                printf("%d%c",res[i*6+j],j==5?'\n':' ');

    }

    return 0;

}