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 0Sample 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 11 0 1 1 0 1
思路分析：
高斯消元法。
一个5*6的矩阵，每个元素都是一个未知量。然后每个位置都对应一个方程，利用高斯消元法解出未知量即可。
ac代码：
#include<stdio.h> #include<iostream> #include<string.h> #include<math.h> #include<algorithm> using namespace std; const int MAXN=40; //有equ个方程，var个变元。增广矩阵行数为equ,分别为0到equ-1,列数为var+1,分别为0到var. int equ,var; int a[MAXN][MAXN];//增广矩阵 int x[MAXN];//解集 int Gauss() { int i,j,k; int max_r; int col; int temp; col=0; for(k=0;k<equ && col<var; k++,col++) //将矩阵化成阶梯型。 { max_r=k; for(i=k+1; i<equ; i++) //找出该列 不为0的那一行。 { if(abs(a[i][col])>abs(a[max_r][col])) max_r=i; } if(max_r!=k) //与第1行交换。 { for(j=col; j<var+1; j++) swap(a[k][j],a[max_r][j]); } if(a[k][col]==0) //该列全为0.则遍历下一行。 { k--; continue; } for(i=k+1; i<equ; i++) { if(a[i][col]!=0) { for(j=col; j<var+1; j++) a[i][j]^=a[k][j]; } } } for(i=var-1; i>=0; i--) { x[i]=a[i][var]; for(j=i+1; j<var; j++) x[i]^=(a[i][j]&&x[j]); } return 0; } void init() //预处理得到 系数矩阵。 { memset(a,0,sizeof(a)); equ=30; var=30; 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; } } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int T; scanf("%d",&T); int iCase=0; while(T--) { iCase++; init(); for(int i=0; i<30; i++) scanf("%d",&a[i][30]); Gauss(); printf("PUZZLE #%d",iCase); for(int i=0; i<30; i++) { if(i%6==0)printf("\n%d",x[i]); else printf(" %d",x[i]); } printf("\n"); } return 0; }