POJ 1222 Gaussian elimination method to solve off-off problem

#include <cstdio> #include <cstring> #include <algorithm>const int maxn = 30;int INP[MAXN + 5][MAXN + 5],  Kase, T;void Gauss () {for (int i = 0; i < MAXN; i++) {int k = I;while (k < MAXN &&!inp[k][i]) k++;for (int j = 0; J <= Maxn; J + +) Std::swap (Inp[i][j], inp[k][j]); for (int j = 0; J < Maxn; J + +) if (i! = J && Inp[j][i]) for (int p = 0; p &lt ; = MAXN; p++) Inp[j][p] = Inp[i][p] ^ inp[j][p];}} int main (int argc, char const *argv[]) {scanf ("%d", &t), while (t--) {memset (INP, 0, sizeof (INP)); for (int i = 0; i < MAXN; i++) scanf ("%d", &AMP;INP[I][MAXN]), inp[i][i] = 1;for (int i = 0; i < MAXN; i++) {if (i% 6! = 0) Inp[i-1][i] = 1;if ( I% 6 = 5) Inp[i + 1][i] = 1;if (i > 5) inp[i-6][i] = 1;if (i <) Inp[i + 6][i] = 1;} Gauss ();p rintf ("PUZZLE #%d\n", ++kase), for (int i = 0; i < MAXN; i++) printf ("%d%c", INP[I][MAXN], I% 6 = 5? ' \ n ': ');} return 0;}

Upstairs for the simple version, downstairs for the version of the template.

#include <cstdio> #include <cstring> #include <algorithm>const int maxn = 30;int INP[MAXN + 5][MAXN + 5], Kase, t;//void Gauss ()//{//for (int i = 0; i < MAXN; i++)//{//int k = i;//while (k < MAXN &&!inp[k][i  ] k++;//for (int j = 0; J <= Maxn; j + +)/Std::swap (Inp[i][j], inp[k][j]);/for (int j = 0; J < Maxn; J + +)/if (i  ! = J && Inp[j][i])//for (int p = 0; p <= maxn; p++)/inp[j][p] = inp[i][p] ^ inp[j][p];//}//}int equ, Var,  X[MAXN], Free_num, Free_x[maxn];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 (Inp[i][col]) > abs (Inp[max_r][col])) Max_r = I;if (INP [Max_r] [col] = = 0) {k--; free_x[free_num++] = col;continue;} if (max_r! = k) for (int j = col; J < var + 1; j + +) Std::swap (Inp[k][j], inp[max_r][j]); for (int i = k + 1; i < equ; I + +) if (Inp[i][col]! = 0) for (int j = col; J < var + 1; j + +) Inp[i][j] ^= INP[K][J];} for (int i = k; i < equ; i++) if (inp[i][col]! = 0) return-1;if (K < Var) return var-k;for (int i = var-1; i > = 0; i--) {X[i] = inp[i][var];for (int j = i + 1; j < var; j + +) X[i] ^= (Inp[i][j] && x[j]);} return 0;} int main (int argc, char const *argv[]) {scanf ("%d", &t), while (t--) {memset (INP, 0, sizeof (INP)); for (int i = 0; i < MAXN; i++) scanf ("%d", &AMP;INP[I][MAXN]), inp[i][i] = 1;for (int i = 0; i < MAXN; i++) {if (i% 6! = 0) Inp[i-1][i] = 1;if ( I% 6 = 5) Inp[i + 1][i] = 1;if (i > 5) inp[i-6][i] = 1;if (i <) Inp[i + 6][i] = 1;}  Equ = $, var = 30;gauss ();p rintf ("PUZZLE #%d\n", ++kase); for (int i = 0; i < MAXN; i++) printf ("%d%c", x[i], I% 6 = 5 ? ' \ n ': ');} return 0;}

Give you a 5 row 6 columns of the matrix represents 30 lights, the matrix inp[i][j]=1 indicates that the light is on, =0 that the light is not lit. Set solution for Press[][]:press[i][j]=1, =0 means no press. So that the final state for all lights is off. Through analysis, we can know: the Order of the buttons do not matter, each button at most only need to press the next time, multi-press Cancel.

When the press acts on the INP, the effect is the corresponding bit modulus 2 plus (XOR), then each point as an unknown, his upper and lower values will determine his value.

POJ 1222 Gaussian elimination method to solve off-off problem