poj 1222 EXTENDED LIGHTS OUT（高斯消元）

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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

Output

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

高斯消元的简单题：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;

int a[35][35],x[35];
int equ,var;//方程数和变元数

void Debug()
{
    for(int i=0;i<equ;i++)
    {
        for(int j=0;j<var;j++)
            printf("%d ",a[i][j]);
        puts("");
    }
}

void gauss()
{
    int max_r;
    for(int i=0;i<equ;i++)
    {
        max_r=i;
        for(int j=i+1;j<equ;j++)
        {
            if(a[j][i]>a[max_r][i]) max_r=j;
        }
        if(max_r!=i)
        {
            for(int k=0;k<=var;k++)
            {
                swap(a[i][k],a[max_r][k]);
            }
        }

        for(int j=0;j<equ;j++)
        if(i!=j&&a[j][i])
        {
            for(int k=i;k<=var;k++)
            a[j][k]=a[i][k]^a[j][k];
        }
    }

    //Debug();

    for(int i=equ-1;i>=0;i--)
    {
        int temp=a[i][var];
        for(int j=i+1;j<var;j++)
        {
            temp-=a[i][j]*x[j];
        }
        x[i]=temp/a[i][i];
    }
}


void init()
{
    memset(a,0,sizeof(a));
    memset(x,0,sizeof(x));
    for(int i=0;i<5;i++)
    for(int j=0;j<6;j++)
    {
        if(i!=0) a[i*6+j][(i-1)*6+j]=1;
        if(i!=4) a[i*6+j][(i+1)*6+j]=1;
        if(j!=0) 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;
    }
}
int main()
{
    int cas;
    equ=var=30;
    scanf("%d",&cas);
    for(int ca=1;ca<=cas;ca++)
    {
        init();
        for(int i=0;i<equ;i++)
        {
            scanf("%d",&a[i][var]);
        }

        gauss();
        printf("PUZZLE #%d\n",ca);
        for(int i=0;i<equ;i++)
        {
            printf("%d",x[i]);
            if(i%6==5) puts("");
            else printf(" ");
        }
    }
    return 0;
}

poj 1222 EXTENDED LIGHTS OUT（高斯消元）

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原文地址：http://blog.csdn.net/knight_kaka/article/details/38797087