USACO Party Lamps

题目大意:一排灯有n个,有４种开关,每种开关能改变一些灯现在的状态(亮的变暗,暗的变亮)现在已知一些灯的亮暗情况,问所以可能的情况是哪些

思路:同一种开关开两次显然是没效果的,那么枚举每个开关是否开就好了，还是暴力大法好

/*{
ID:a4298442
PROB:lamps
LANG:C++
}
*/
#include<iostream>
#include<fstream>
#include<cstring>
#include<algorithm>
#define maxn 109
using namespace std;
ifstream fin("lamps.in");
ofstream fout("lamps.out");
//#define fin cin
//#define fout cout
int lamp[maxn],temp[maxn],n1,n2,a[maxn],b[maxn];
struct T
{
char ch[maxn];
}ans[maxn];
void but1(int n)
{
for(int i=1;i<=n;i++)lamp[i]^=1;
}
void but2(int n)
{
for(int i=1;i<=n;i+=2)lamp[i]^=1;
}
void but3(int n)
{
for(int i=2;i<=n;i+=2)lamp[i]^=1;
}
void but4(int n)
{
for(int i=0;i*3+1<=n;i++)lamp[i*3+1]^=1;
}
int check(int n,int c,int count_now)
{
for(int i=1;i<n1;i++)if(lamp[a[i]]==0)return 0;
for(int i=1;i<n2;i++)if(lamp[b[i]]==1)return 0;
if(count_now>c)return 0;
int u=c-count_now;
if(u&1)return 0;
return 1;
}
int cmp(T x, T y)
{
return strcmp(x.ch+1,y.ch+1)>=0?false:true;
}
int main()
{
int n,c;
fin>>n>>c;
while(fin>>a[++n1]&&a[n1]!=-1);
while(fin>>b[++n2]&&b[n2]!=-1);
for(int i=1;i<=n;i++)temp[i]=1;
int h=0;
for(int i=0;i<=(1LL<<4)-1;i++)
{
int count_now=0;
memcpy(lamp,temp,sizeof(lamp));
for(int j=1,idx=1;j<=i;j<<=1,idx++)if((i&j)!=0 )
{
count_now++;
if(idx==1)but1(n);
if(idx==2)but2(n);
if(idx==3)but3(n);
if(idx==4)but4(n);
}
if(check(n,c,count_now))
{
h++;
for(int i=1;i<=n;i++)
{
ans[h].ch[i]=lamp[i]+'0';
}
}
}
if(h==0)fout<<"IMPOSSIBLE"<<endl;else
{
sort(ans+1,ans+1+h,cmp);
for(int i=1;i<=h;i++)fout<<ans[i].ch+1<<endl;
}
return 0;
}