poj 1222 EXTENDED LIGHTS OUT 【高斯消元】

EXTENDED LIGHTS OUT

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7817		Accepted: 5087

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

醉了，学校大二下学期才开线代。花了那么长时间就只为了看概念。。。o(╯□╰)o  

明天补上模板。顺便ORZ大bin神。

题意：有一个5*6的矩阵，每个元素代表1盏灯，0表示灯关，1表示灯开。若你改变了第a[i][j]盏灯的状态，相应地还会改变它四周a[i-1][j]、a[i][j-1]、a[i+1][j]、a[i][j+1]这些灯的状态。因为同一盏灯连续按两次又回到初始状态，所以每个灯只需按一次，若选择按第a[i][j]盏灯，则x[i][j] = 1，否则为0。现在给你这些灯的初始状态，让你找出一种方案x[][]使得所有灯全熄灭。

高斯消元——解n元一次线性方程组。

1，将方程组化简成增广矩阵A；

2，利用列主元法 + 加减消元法 进行消元，将矩阵A变成阶梯矩阵；

3，在解唯一的情况下，形成严格的上三角矩阵，然后就是回代过程——自底向上求所有变量的解。

思路：入门题目吧，构建的矩阵比较简单。

对每盏灯，列一个异或方程，把可能影响它的灯和自己置1，其余灯置0。

比如说灯初始状态为a[i][j]，全灭的方案为x[i][j]。

则有——

a[i][j] ^ x[i-1][j]*1 ^ x[i][j-1]*1 ^ x[i+1][j]*1 ^ x[i][j+1]*1 ^ x[i][j]*1 ^ (其余不相关的灯*0) = 0。

两边同时异或a[i][j]得

x[i-1][j]*1 ^ x[i][j-1]*1 ^ x[i+1][j]*1 ^ x[i][j+1]*1 ^ x[i][j]*1 ^ (其余不相关的灯*0) = a[i][j]。

30个方程组，30个变元，然后直接模板KO。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[31][31];//增广矩阵
int equ, var;//equ个方程 var个变元
int x[31];//解
void init_A()
{
    memset(a, 0, sizeof(a));
    equ = 30, var = 30;//30个方程 30个变元
    for(int i = 0; i < 5; i++)
    {
        for(int j = 0; j < 6; j++)
        {
            a[i*6+j][i*6+j] = 1;
            if(i > 0) a[i*6+j][(i-1)*6+j] = 1;
            if(j > 0) a[i*6+j][i*6+j-1] = 1;
            if(i < 4) a[i*6+j][(i+1)*6+j] = 1;
            if(j < 5) a[i*6+j][i*6+j+1] = 1;
        }
    }
    for(int i = 0; i < 30; i++)
        scanf("%d", &a[i][30]);
}
int Gauss()
{
    int k, max_r;
    int col = 0;
    for(k = 0; k < equ && col < var; k++, col++)
    {
        //列主元法
        //找第col元素绝对值最大的行i(i > k) 与 当前行交换
        max_r = k;
        for(int i = k + 1; i < equ; i++)
            if(a[i][col] > a[k][col])
                max_r = i;
        if(max_r != k)//找到 —— 交换
        {
            for(int i = col; i < var+1; i++)
                swap(a[k][i], a[max_r][i]);
        }
        if(a[k][col] == 0)//k行下面的元素全为0
        {
            k--;
            continue;
        }
        //加减消元
        for(int i = k + 1; i < equ; i++)
        {
            if(a[i][col] != 0)
            {
                for(int j = col; j < var+1; j++)
                    a[i][j] ^= a[k][j];
            }
        }
    }
    //根据上三角矩阵 回代求解
    for(int i = var-1; i >= 0; i--)
    {
        x[i] = a[i][var];
        for(int j = i + 1; j < var; j++)
            x[i] ^= (a[i][j] * x[j]);
    }
    return 0;
}
int main()
{
    int t, k = 1;
    scanf("%d", &t);
    while(t--)
    {
        init_A();
        Gauss();
        printf("PUZZLE #%d\n", k++);
        for(int i = 0; i < 30; i++)
        {
            if((i+1) % 6 == 0)
                printf("%d\n", x[i]);
            else
                printf("%d ", x[i]);
        }
    }
    return 0;
}