POJ 1222 EXTENDED LIGHTS OUT
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题目分析
读完整个题目一直感觉会有多解,但是题目好像并没有这个意思,于是我就按有唯一解的写法写的。直接构造一个30个方程,然后高斯消元就行了。
#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 35;typedef int Matrix[maxn][maxn];int dir[5][2] = {{1, 0}, {-1, 0},{0, 0},{0, -1},{0, 1}};Matrix A;void reset(){ for(int i = 0; i < 5; i++) for(int j = 0; j < 6; j++) for(int k = 0; k < 5; k++){ int x = i + dir[k][0]; int y = j + dir[k][1]; if(x >= 0 && x < 5 && y >= 0 && y < 6) A[i*6+j][x*6+y] = 1; }}void guess(){ for(int i = 0; i < 30; i++){ int k; for(k = i; k < 30; k++) if(A[k][i]) break; for(int j = 0; j <= 30; j++) swap(A[k][j], A[i][j]); for(int j = 0; j < 30; j++) if(j != i && A[j][i]) for(int k = 0; k <= 30; k++) A[j][k] = A[j][k]^A[i][k]; }}int main(){ int T; scanf("%d", &T); for(int kase = 1; kase <= T; kase++){ reset(); for(int i = 0; i < 30; i++) scanf("%d", &A[i][30]); printf("PUZZLE #%d\n", kase); guess(); for(int i = 0; i < 30; i++){ if(i%6 == 5) printf("%d\n", A[i][30]); else printf("%d ", A[i][30]); } } return 0;}
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