https://www.acwing.com/problem/content/description/293/
Ideas:
A very important point is that after placing the legal horizontal blocks, the placement of the vertical blocks is determined.
So enumerate the status of each column, expressed in binary. j is the state of the previous column, and k is the state of the column.
It can be known that j&k==0 and the number of consecutive 0s in each segment of j|k is even.
Therefore, the state of conformity where the number of consecutive 0s in each segment of j|k is an even number is consistent in each column state. Just pre-process in advance.
if( (j&k)==0 &&(st[j|k] )
dp[i][j]+=dp[i-1][k]
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=14;
typedef int LL;
inline LL read(){LL x=0,f=1;char ch=getchar(); while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
long long dp[maxn][1<<maxn];
bool st[1<<maxn];
int main(void)
{
LL n,m;
while(scanf("%d%d",&n,&m)&&n!=0&&m!=0){
for(LL i=0;i<(1<<n);i++){
st[i]=true;
LL cnt=0;///记录1前的连续0的个数
for(LL j=0;j<n;j++){
if(i&(1<<j)){
if(cnt&1){
st[i]=false;break;
}
cnt=0;
}
else cnt++;
}
if(cnt&1) st[i]=false;///下面只有0的
}
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(LL i=1;i<=m;i++){///枚举每一列
for(LL j=0;j<(1<<n);j++){
for(LL k=0;k<(1<<n);k++){
if((j&k)==0&&(st[j|k]==true)){
dp[i][j]+=dp[i-1][k];
}
}
}
}
printf("%lld\n",dp[m][0]);
}
return 0;
}