用dp[i][j]表示要贴第i层时，第i-1层的状态， 然后每一层dfs枚举一下情况即可
dfs(x, y, pre, now) x层数 y枚举的当前列 pre 前一层状态, now当前层状态
根据当前列y的这位上pre和now状态，确定要贴的图形， 注意贴完以后pre层必须满状态，在贴的过程中处理必须优先把pre层贴满。
状态转移如下图：.表示空，*表示满
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define LL long long
int n, m;
LL dp[11][1<<9];
int f(int a) {
return 1<<a;
}
LL cnt;
int N;
void dfs(int x, int y, int pre, int now) {
if(y >= m) {
dp[x+1][now] += cnt;
return;
}
if( pre&f(y) && now&f(y)) {
dfs(x, y+1, pre, now);
return;
}
if( !(pre&f(y)) && !(now&f(y)) ) {
dfs(x, y+1, pre|f(y), now|f(y)); //*
if(y+1 < m) { //*
if( !(now&f(y+1)) ) //*
dfs(x, y+1, pre|f(y), now|f(y)|f(y+1)); //**
if( !(pre&f(y+1)) ) //**
dfs(x, y+1, pre|f(y)|f(y+1), now|f(y)); //*
if( !(pre&f(y+1)) && !(now&f(y+1)) ) //**
dfs(x, y+1, pre|f(y)|f(y+1), now|f(y+1)); // *
}
return;
}
if( pre&f(y) && !(now&f(y)) ) {
if( y+1 < m && !(now&f(y+1)) ) //*
dfs(x, y+1, pre|f(y), now|f(y)|f(y+1)); //**
if( y+1 < m && !(pre&f(y+1)) && !(now&f(y+1)) ) // *
dfs(x, y+1, pre|f(y+1), now|f(y)|f(y+1)); //**
dfs(x, y+1, pre, now); // 不放
return;
}
if( !(pre&f(y)) && now&f(y)) {
if(y+1 < m && !(pre&f(y+1) && !(now&f(y+1))) ) //**
dfs(x, y+1, pre|f(y)|f(y+1), now|f(y+1)); // *
}
}
int main() {
int i, j;
while(~scanf("%d%d", &n, &m)) {
N = 1<<m;
memset(dp, 0, sizeof(dp));
dp[0][N-1] = 1;
for(i = 0; i < n; i++)
for(j = 0; j < N; j++) if(dp[i][j]) {
cnt = dp[i][j];
dfs(i, 0, j, 0);
}
cout << dp[n][N-1] << endl;
}
}
/*
2 3
5
9 9
38896105985522272
6 6
5350806
*/