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## 状压DP，有不太详细的注释，719ms

Posted by OZY123 at 2016-11-02 20:09:27 on Problem 2411
```#include<cstdio>
#include<cstring>
const long long N=12;
long long n,m;
long long f[N][(1<<12)];//第i行的状态为j，i-1行都已经全部填满有多少种状态
long long tt;
bool ooo[(1<<12)];
//0:没有覆盖  1:已经覆盖
bool check (long long x,long long y)//x和y是否可以相得益彰
{
if ((x|y)!=tt-1) return false;
long long kk=(x&y);
return ooo[kk];
}
void prepare ()
{
memset(ooo,true,sizeof(ooo));
for (long long u=0;u<(1<<12);u++)
{
long long x=u;
bool last=false;
while (x>0)
{
if ((x&1)==1)
last=!last;
else if (last==true)
{
ooo[u]=false;
break;
}
x=x>>1;
}
if (last==true) ooo[u]=false;
}
return ;
}
int main()
{
prepare();
while (true)
{
memset(f,0,sizeof(f));
scanf("%I64d%I64d",&n,&m);
if (n==0&&m==0) break;
if (((n*m)&1)==1) {printf("0\n");continue;}
tt=(1<<m);
for (long long u=0;u<tt;u++)
if (ooo[u]==true)
f[1][u]=1;
for (long long u=2;u<=n;u++)
for (long long j=0;j<tt;j++)//枚举状态
for (long long k=0;k<tt;k++)//枚举上一行的状态
if (check(j,k)==true)
f[u][j]+=f[u-1][k];
printf("%I64d\n",f[n][(tt-1)]);
}
return 0;
}```

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