题解 | #字符框#

借助掩码将face映射成一个二进制数，如果字符的二进制数相同，则字符相同

n,m=map(int,input().split())
t=0
g=[]
for i in range(n):
    g.append(list(input()))
for x in 'face': t|=1<<(ord(x)-97)
r=0
for i in range(n-1):
    for j in range(m-1):
        cur=0
        for x in [g[i][j],g[i+1][j],g[i][j+1],g[i+1][j+1]]:
            cur |= 1<< ord(x)-97
        if cur==t: r+=1
print(r)