poj 1222 EXTENDED LIGHTS OUT（高斯消元（开关问题 对2取模的01方程组））

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EXTENDED LIGHTS OUT

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7447 Accepted: 4853

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

20 1 1 0 1 01 0 0 1 1 10 0 1 0 0 11 0 0 1 0 10 1 1 1 0 00 0 1 0 1 01 0 1 0 1 10 0 1 0 1 11 0 1 1 0 00 1 0 1 0 0

Sample Output

PUZZLE #11 0 1 0 0 11 1 0 1 0 10 0 1 0 1 11 0 0 1 0 00 1 0 0 0 0PUZZLE #21 0 0 1 1 11 1 0 0 0 00 0 0 1 0 01 1 0 1 0 11 0 1 1 0 1

5*6的矩阵中有灯和开关 0表示灯灭1表示灯亮  按下一个开关 这个开关上下左右的灯的状态就会发生变化

现在要使所有的灯都关闭 问需要按下哪些按钮  答案输入开关的矩阵图

按钮的顺序可以随便  按钮只能按一次

第一次写高斯消元  学习kuangbin的板子

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define MEM(a,x) memset(a,x,sizeof a)#define eps 1e-8#define MOD 10009#define MAXN 40#define MAXM 100010#define INF 99999999#define ll __int64#define bug cout<<"here"<<endl#define fread freopen("ceshi.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)using namespace std;int Read(){    char c = getchar();    while (c < '0' || c > '9') c = getchar();    int x = 0;    while (c >= '0' && c <= '9') {        x = x * 10 + c - '0';        c = getchar();    }    return x;}void Print(int a){     if(a>9)         Print(a/10);     putchar(a%10+'0');}int equ,var;//equ个方程 var个变元 增广矩阵行数为equ 列数为var+1 分别为0到varint a[MAXN][MAXN];//增广矩阵int x[MAXN];//解集int free_x[MAXN];//存储自由变元（多解枚举自由变元可以使用）int free_num;//自由变元的个数void init(){    MEM(a,0); MEM(x,0);    equ=5*6; var=5*6;    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;        }}//返回值为-1表示无解 为0是唯一解 否则返回自由变元个数int Gauss(){    int max_r,col,k;    free_num=0;    for(k=0,col=0;k<equ&&col<var;k++,col++)    {        max_r=k;        for(int i=k+1;i<equ;i++)        {            if(abs(a[i][col])>abs(a[max_r][col]))                max_r=i;        }        if(a[max_r][col]==0)        {            k--;            free_x[free_num++]=col;//这个是自由变元            continue;        }        if(max_r!=k)        {            for(int j=col;j<var+1;j++)                swap(a[k][j],a[max_r][j]);        }        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=k;i<equ;i++)        if(a[i][col]!=0)            return -1;//无解    if(k<var) return var-k;//自由变元个数    //唯一解  回带    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(){//    fread;    int tc;    scanf("%d",&tc);    int cs=1;    while(tc--)    {        init();        for(int i=0;i<30;i++)            scanf("%d",&a[i][30]);        Gauss();        printf("PUZZLE #%d\n",cs++);        for(int i=0;i<5;i++)        {            for(int j=0;j<6;j++)            {                if(j) printf(" ");                printf("%d",x[i*6+j]);            }            puts("");        }    }    return 0;}