Poj 1222 extended lights out (Math _ Gaussian deyuan)

To give a 01 matrix of 5*6, 0 indicates that the light is dark, and 1 indicates that the light is on. Each position in the matrix represents a button. When the button is pressed, the lights around it (up and down) are in the opposite state. Ask how to turn all the lights into a dark one.

Solution: This type of switch problem is a classic Gaussian elimination problem. When doing this, I can think of how to create the matrix, but the solution below does not know how to solve it, the line generation teacher died early.
The final state of each lamp is determined only by itself and the button next to it. There are 30 lamps. The final state of each lamp is 0. If the initial state is 0, the center must be an even number of times; otherwise, it is an odd number of times. Let's set 30 equations. The value modulus 2 on the right side of each equation equals the value corresponding to the matrix. In this way, the values in the matrix are all 0. If some vertices can affect this vertex, then the vertices are marked as 1. In this way, each vertex corresponds to an unknown number, and the last value is 0 or 1.
After the matrix is constructed, you can modify the Gaussian elimination template. Because this question has only 0 and 1 in a special solution, you only need to use a few bits.

Test data:
1
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
Out:
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

C producer code:
[Cpp]
<Span style = "color: # ff0000;" >#include <stdio. h>
# Include <string. h>
# Include <algorithm>
Using namespace std;
# Define MOD 2
# Define MAX 35
 
 
Int n, m, x [MAX];
Int arr [MAX] [MAX];
Int mat [MAX] [MAX];
Int dir [4] [2] = {}, {-}, {}, {0,-1 }};
 
 
Void Initial (){
 
Int I, j, k, row, col;
 
 
N = m = 5*6;
Memset (mat, 0, sizeof (mat ));
For (I = 0; I <5; ++ I)
For (j = 0; j <6; ++ j ){
 
Row = I * 6 + j;
Mat [row] [row] = 1;
Mat [row] [m] = arr [I] [j];
For (k = 0; k <4; ++ k ){
 
Int ii = dir [k] [0] + I;
Int jj = dir [k] [1] + j;
If (ii> = 0 & ii <5
& Jj> = 0 & jj <6)
Mat [row] [ii * 6 + jj] = 1;
}
}
}
Int Gcd (int x, int y ){
 
Int r = x % y;
While (r ){
 
X = y, y = r;
R = x % y;
}
Return y;
}
Int Lcm (int x, int y ){
 
Return x/Gcd (x, y) * y;
}
Void Gauss_AC (){
 
Int I, j, row, max_r, col;
Int ta, tb, temp, LCM;
 
 
Row = col = 0;
For (; row <n & col <m; ++ row, ++ col ){
 
Max_r = row;
For (j = row + 1; j <n; ++ j)
If (mat [j] [col]) {
 
Max_r = j;
Break;
}
If (mat [max_r] [col] = 0 ){
 
Row --;
Continue;
}
 
 
If (row! = Max_r)
For (j = col; j <= m; ++ j)
Swap (mat [row] [j], mat [max_r] [j]);
For (I = row + 1; I <n; ++ I) // the first element of each row is reduced to 0. Therefore, use the row to remove an exclusive or I row.
If (mat [I] [col]) for (j = col; j <= m; ++ j)
Mat [I] [j] ^ = mat [row] [j];
}
 
 
For (I = m-1; I> = 0; -- I ){
 
X [I] = mat [I] [m]; // mat [I] [I] is 1, the sum of mat [I] [m] and other columns must be 0, so we use an exclusive or. If the exclusive or is 1, x [I] is 1.
For (j = I + 1; j <m; ++ j)
X [I] ^ = (mat [I] [j] & x [j]);
}
}
 
 
Int main ()
{
Int I, j, k, t, cas = 0;
 
 
Scanf ("% d", & t );
While (t --){
 
For (I = 0; I <5; ++ I)
For (j = 0; j <6; ++ j)
Scanf ("% d", & arr [I] [j]);
 
 
Initial ();
Gauss_AC ();
Printf ("PUZZLE # % d \ n", ++ cas );
For (I = 0; I <m; ++ I)
Printf ("% d % c", x [I], (I % 6) = 5? '\ N ':'');
}
}
 
</Span>

Author: woshi250hua