poj1830 开关问题

记
\(a_{i,j}\) 表示第 \(j\) 个开关对第 \(i\) 号开关产生的影响，\(x_i\) 为对第 \(i\) 个开关的操作，则

\[\begin{cases} a_{1,1}x_1\ \mathrm{xor}\ a_{1,2}x_2\ \mathrm{xor}\ \cdots\ \mathrm{xor}\ a_{1,n}x_n=start_1 \ \mathrm{xor}\ end_1 \\ a_{2,1}x_1\ \mathrm{xor}\ a_{2,2}x_2\ \mathrm{xor}\ \cdots\ \mathrm{xor}\ a_{2,n}x_n=start_2 \ \mathrm{xor}\ end_2 \\ \cdots \\ a_{n,1}x_1\ \mathrm{xor}\ a_{n,2}x_2\ \mathrm{xor}\ \cdots\ \mathrm{xor}\ a_{n,n}x_n=start_n \ \mathrm{xor}\ end_n \\ \end{cases}\]

解就是了

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int T, a[35], sta, end, n, uu, vv;
int gauss(){
	for(int i=1; i<=n; i++){
		int maxi=i;
		for(int j=i+1; j<=n; j++)
			if(a[j]>a[maxi])
				maxi = j;
		swap(a[maxi], a[i]);
		if(a[i]==0)	return 1<<(n-i+1);
		if(a[i]==1)	return -1;
		for(int j=n; j; j--)
			if(a[i]&(1<<j)){
				for(int k=1; k<=n; k++)
					if(k!=i && a[k]&(1<<j))
						a[k] ^= a[i];
				break;
			}
	}
	return 1;
}
int main(){
	cin>>T;
	while(T--){
		memset(a, 0, sizeof(a));
		sta = end = 0;
		scanf("%d", &n);
		for(int i=1; i<=n; i++){
			scanf("%d", &uu);
			a[i] ^= uu;
		}
		for(int i=1; i<=n; i++){
			scanf("%d", &uu);
			a[i] ^= uu;
			a[i] |= 1<<i;
		}
		while(scanf("%d %d", &uu, &vv)!=EOF){
			if(!uu && !vv)	break;
			a[vv] |= 1<<uu;
		}
		int re=gauss();
		if(re<0)	printf("Oh,it's impossible~!!\n");
		else	printf("%d\n", re);
	}
	return 0;
}