高斯消元求异或方程组。
每个灯连续操作两次等于没有操作,所以每盏灯只有0/1的操作状态,记为(x_i)
第(i)盏灯对第(j)盏灯有影响,则(a_{j,i} = 1),反之(a_{j,i} = 0)
第(i)盏灯初末状态不一样,(a_{i,n+1} = 1)反之为(0)
然后我们得到矩阵
[left[
egin{array}{ccc}
a_{1,1} & a_{1,2} & ldots & a_{1,n} & | & a_{1,n + 1} \\
a_{2,1} & a_{2,2} & ldots & a_{2,n} & | & a_{2,n + 1} \\
vdots & vdots & vdots & vdots & | & vdots \\
a_{n,1} & a_{n,2} & ldots & a_{n,n} & | & a_{1,n + 1} \\
end{array}
ight]
]
直接高斯消元,用异或消去其他位上的(1)
如果出现第(j)列上所有的(1)都消去了,那么第(j)盏灯操不操作都可以,我们称为一个自由元。
统计自由元的个数(k),(Ans = 2^k)
如果第(i)行(a_{i,1} ~ a_{i,n})均为(0),但是(a_{i,n+1} = 1)则无解。
以上
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int n;
int a[35],b[35];
int g[35][35];
long long ksm(long long x,int y){
long long z = 1;
while(y){
if(y & 1) z = z * x;
y >>= 1;
x = x * x;
}
return z;
}
int main(){
int T; scanf("%d",&T);
while(T --){
memset(g,0,sizeof(g));
scanf("%d",&n);
for(int i = 1; i <= n; ++ i) scanf("%d",&a[i]);
for(int i = 1; i <= n; ++ i) scanf("%d",&b[i]);
for(int i = 1; i <= n; ++ i) if(a[i] == b[i]) g[i][n + 1] = 0; else g[i][n + 1] = 1;
int x,y;
while(1){
scanf("%d%d",&x,&y);
if(x == 0 && y == 0) break;
g[y][x] = 1;
}
for(int i = 1; i <= n; ++ i) g[i][i] = 1;
int now = 1;
for(int i = 1; i <= n; ++ i){
bool flag = 0; int pos = -1;
for(int j = now; j <= n; ++ j){
if(g[j][i] == 1) { flag = 1; pos = j; break; }
}
if(!flag) continue;
for(int j = 1; j <= n + 1; ++ j) swap(g[now][j],g[pos][j]);
for(int j = now + 1; j <= n; ++ j){
if(g[j][i] == 0) continue;
for(int k = 1; k <= n + 1; ++ k)
g[j][k] ^= g[now][k];
}
++ now;
}
bool flag = 1;
for(int i = now; i <= n; ++ i){
if(g[i][n + 1] != 0) { flag = 0; break; }
}
if(!flag) puts("Oh,it's impossible~!!");
else{
printf("%lld
",ksm(2, n - now + 1));
}
}
return 0;
}