POJ1830（异或高斯消元）

对于某个开关，都有n个选项可能影响它的结果，如果会影响，则系数为1，否则系数为0；最后得到自由元的个数，自由元可选0也可选1.

#include <cstdio>
#include <algorithm>

int T, n, a[30], x, y;

int gauss() {
	for (int i = 1; i <= n; i++) {
		//列主
		for (int j = i + 1; j <= n; j++) {
			if (a[j] > a[i]) {
				std::swap(a[i], a[j]);
			}
		}
		if (a[i] == 0)	return 1 << (n - i + 1);
		if (a[i] == 1)	return -1;
		//消元
		for (int k = n; k; k--) {
			if (a[i] & (1 << k)) {
				for (int j = 1; j <= n; j++) {
					if (i != j && a[j] & (1 << k)) {
						a[j] ^= a[i];
					}
				}
				break;
			}
		}
	}
	return 1;
}

int main(int argc, char const *argv[]) {
	scanf("%d", &T);
	while (T--) {
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) {
			scanf("%d", &a[i]);	
		}
		for (int i = 1, j; i <= n; i++) {
			scanf("%d", &j);
			a[i] ^= j;//等号右侧
			a[i] |= 1 << i;//a[i][i] = 1
		}
		while (~scanf("%d %d", &x, &y) && (x | y)) {
			a[y] |= 1 << x;//a[y][x] = 1
		}
		int ans = gauss();
		if (ans > 0)	printf("%d
", ans);
		else	puts("Oh,it's impossible~!!");
	}
	return 0;
}