POJ1222EXTENDED LIGHTS OUT（高斯消元）

EXTENDED LIGHTS OUT

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11815		Accepted: 7577

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

/*
* @Author: LyuC
* @Date:   2017-10-11 10:53:15
* @Last Modified by:   LyuC
* @Last Modified time: 2017-10-11 19:45:51
*/
/*
 题意：给你一个5*6的矩阵状态，每次你可以反转一个格子，上下左右的一个格子，会同时被反转
    问你怎么样反转才能到达给定的状态

 思路：以前用状压做过，最近看高斯消元所以再做一遍，先抽象出来线性方程：
    (a[1]&&x[1])^(a[2]&&x[2])^...(a[n]&&x[n])=b[1]：
        a[i]就是a[i][j]状态数组，x[i]是解集，表示的是每个位置按或者不按，b[i]就是i这个位置
        的状态；
    然后高斯消元解出x数组就可以了
*/
#include <iostream>
#include <stdio.h>
#include <cmath>
#include <string.h>

#define MAXN 35

using namespace std;

int t;
int x[MAXN];//解集
int a[MAXN][MAXN];
int equ,var;

inline void Guass(){
    for(int i=1;i<=var+1;i++){
        x[i]=0;//初始化解
    }
    for(int col=1,k=1;col<=var&&k<=equ;col++,k++){//k当前枚举到第几行，col当前枚举到第几列
        //找系数最大行
        int maxk=k;
        for(int i=k+1;i<=equ;i++){
            if(a[maxk][col]<a[i][col]){
                maxk=i;
            }
        }
        if(maxk!=k){
            for(int i=col;i<=var+1;i++){
                swap(a[k][i],a[maxk][i]);
            }
        }
        if(a[k][col]==0){
            k--;
            continue;
        }
        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];
                }
            }
        }
    }
    for(int i=var;i>=1;i--){
        x[i]=a[i][var+1];
        for(int j=i+1;j<=var;j++){
            x[i]^=(a[i][j]&&x[j]);
        }
    }
}

inline void init(){//初始化矩阵
    memset(a,0,sizeof a);
    equ=30;
    var=30;
    for(int i=1;i<=5;i++){
        for(int j=1;j<=6;j++){
            int t=(i-1)*6+j;
            a[t][t]=1;
            if(i>1) a[(i-2)*6+j][t]=1;
            if(i<5) a[i*6+j][t]=1;
            if(j>1) a[(i-1)*6+j-1][t]=1;
            if(j<6) a[(i-1)*6+j+1][t]=1;
        }
    }
}

int main(){ 
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++){
        printf("PUZZLE #%d
",ca);
        init();
        for(int i=1;i<=30;i++){
            scanf("%d",&a[i][31]);
        }
        Guass();
        for(int i=1;i<=30;i++){
            if(i%6==0){
                printf("%d
",x[i]);
            }else{
                printf("%d ",x[i]);
            }
        }
    }
    return 0;
}