首先特判全部都是A或者全部都是B或者$n=1$的情况。
然后把矩阵四周都填充上A，枚举一个块，分以下情况讨论：
1.在它四周选两个块扩展，此时平方暴力枚举即可。
2.在它四周选定一个方向扩展两步。
3.选择一个角落，斜着扩展一步，再扩展另一步。
时间复杂度$O(n^2)$。
#include<cstdio>
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define N 1010
int n,m,i,j,x,y,l,r,f[N][N],ans,t,q[10000][3],cnt;bool a[N][N];char ch[N];
struct P{int x,y,l,r,s;}s[N*N];
inline void up(int b){if(ans<b)ans=b;}
inline void add(int x,int y,int z){
if(x&&x==y)return;
q[++cnt][0]=x,q[cnt][1]=y,q[cnt][2]=z;
}
int main(){
scanf("%d",&n);gets(ch);
if(n==1)return printf("1",ans),0;
rep(i,1,n)for(gets(ch+1),j=1;j<=n;j++)if(ch[j]=='B')a[i][j]=1;
rep(i,1,n)rep(j,1,n)if(a[i][j]&&!f[i][j]){
for(x=i;a[x][j];x++);
for(y=j;a[i][y];y++);
s[++m].x=i,s[m].y=--x,s[m].l=j,s[m].r=--y;
up(s[m].s=(x-i+1)*(y-j+1));
rep(I,i,x)rep(J,j,y)f[I][J]=m;
}
if(ans==n*n)return printf("%d",ans),0;
for(i=n;~i;i--)for(j=n;~j;j--)f[i+1][j+1]=f[i][j];
for(i=1;i<=m;i++)s[i].x++,s[i].y++,s[i].l++,s[i].r++;
n+=2;
ans=2;
rep(i,1,m){
x=s[i].x,y=s[i].y,l=s[i].l,r=s[i].r,cnt=0;
if(x>2)rep(j,l,r){
if(f[x-2][j])add(f[x-2][j],0,0);
else{
t=s[i].s+2+s[f[x-2][j-1]].s+s[f[x-2][j+1]].s;
if(f[x-2][j-1]!=f[x-1][j-1])t+=s[f[x-1][j-1]].s;
if(f[x-2][j+1]!=f[x-1][j+1])t+=s[f[x-1][j+1]].s;
if(x>3)t+=s[f[x-3][j]].s;
up(t);
}
}
if(y+2<=n)rep(j,l,r){
if(f[y+2][j])add(f[y+2][j],0,0);
else{
t=s[i].s+2+s[f[y+2][j-1]].s+s[f[y+2][j+1]].s;
if(f[y+2][j-1]!=f[y+1][j-1])t+=s[f[y+1][j-1]].s;
if(f[y+2][j+1]!=f[y+1][j+1])t+=s[f[y+1][j+1]].s;
if(y+3<=n)t+=s[f[y+3][j]].s;
up(t);
}
}
if(l>2)rep(j,x,y){
if(f[j][l-2])add(f[j][l-2],0,0);
else{
t=s[i].s+2+s[f[j-1][l-2]].s+s[f[j+1][l-2]].s;
if(f[j-1][l-2]!=f[j-1][l-1])t+=s[f[j-1][l-1]].s;
if(f[j+1][l-2]!=f[j+1][l-1])t+=s[f[j+1][l-1]].s;
if(l>3)t+=s[f[j][l-3]].s;
up(t);
}
}
if(r+2<=n)rep(j,x,y){
if(f[j][r+2])add(f[j][r+2],0,0);
else{
t=s[i].s+2+s[f[j-1][r+2]].s+s[f[j+1][r+2]].s;
if(f[j-1][r+2]!=f[j-1][r+1])t+=s[f[j-1][r+1]].s;
if(f[j+1][r+2]!=f[j+1][r+1])t+=s[f[j+1][r+1]].s;
if(r+3<=n)t+=s[f[j][r+3]].s;
up(t);
}
}
if(f[x-1][l-1]){
if(x>2)add(f[x-1][l-1],f[x-2][l],0);
if(l>2)add(f[x-1][l-1],f[x][l-2],0);
}
if(f[y+1][l-1]){
if(y+2<=n)add(f[y+1][l-1],f[y+2][l],0);
if(l>2)add(f[y+1][l-1],f[y][l-2],0);
}
if(f[x-1][r+1]){
if(x>2)add(f[x-1][r+1],f[x-2][r],0);
if(r+2<=n)add(f[x-1][r+1],f[x][r+2],0);
}
if(f[y+1][r+1]){
if(y+2<=n)add(f[y+1][r+1],f[y+2][r],0);
if(r+2<=n)add(f[y+1][r+1],f[y][r+2],0);
}
if(x==y){
if(l>2)add(f[x][l-2],f[x-1][l-1],f[x+1][l-1]);
if(r+2<=n)add(f[x][r+2],f[x-1][r+1],f[x+1][r+1]);
}
if(l==r){
if(x>2)add(f[x-2][l],f[x-1][l-1],f[x-1][l+1]);
if(y+2<=n)add(f[y+2][l],f[y+1][l-1],f[y+1][l+1]);
}
up(s[i].s+2);
rep(j,1,cnt){
up(s[i].s+2+s[q[j][0]].s+s[q[j][1]].s+s[q[j][2]].s);
rep(k,j+1,cnt){
t=s[i].s+2+s[q[j][0]].s+s[q[j][1]].s+s[q[j][2]].s+s[q[k][0]].s+s[q[k][1]].s+s[q[k][2]].s;
if(q[k][0]==q[j][0]||q[k][0]==q[j][1]||q[k][0]==q[j][2])t-=s[q[k][0]].s;
if(q[k][1]==q[j][0]||q[k][1]==q[j][1]||q[k][1]==q[j][2])t-=s[q[k][1]].s;
if(q[k][2]==q[j][0]||q[k][2]==q[j][1]||q[k][2]==q[j][2])t-=s[q[k][2]].s;
up(t);
}
}
if(x>1&&l>1&&!f[x-1][l-1]){
t=s[i].s+2+s[f[x-2][l-1]].s+s[f[x-1][l-2]].s;
if(x==y&&l==r){
if(f[x-1][l-2]!=f[x][l-2])up(t+s[f[x][l-2]].s+s[f[x+1][l-1]].s);else up(t+s[f[x+1][l-1]].s);
if(f[x-2][l-1]!=f[x-2][l])up(t+s[f[x-2][l]].s+s[f[x-1][l+1]].s);else up(t+s[f[x-1][l+1]].s);
}else{
if(f[x-1][l-2]!=f[x][l-2])up(t+s[f[x][l-2]].s);
if(f[x-2][l-1]!=f[x-2][l])up(t+s[f[x-2][l]].s);
}
}
if(x>1&&r<n&&!f[x-1][r+1]){
t=s[i].s+2+s[f[x-2][r+1]].s+s[f[x-1][r+2]].s;
if(x==y&&l==r){
if(f[x-1][r+2]!=f[x][r+2])up(t+s[f[x][r+2]].s+s[f[x+1][l+1]].s);else up(t+s[f[x+1][l+1]].s);
if(f[x-2][r+1]!=f[x-2][r])up(t+s[f[x-2][r]].s+s[f[x-1][l-1]].s);else up(t+s[f[x-1][l-1]].s);
}else{
if(f[x-1][r+2]!=f[x][r+2])up(t+s[f[x][r+2]].s);
if(f[x-2][r+1]!=f[x-2][r])up(t+s[f[x-2][r]].s);
}
}
if(y<n&&l>1&&!f[y+1][l-1]){
t=s[i].s+2+s[f[y+2][l-1]].s+s[f[y+1][l-2]].s;
if(x==y&&l==r){
if(f[y+1][l-2]!=f[y][l-2])up(t+s[f[y][l-2]].s+s[f[x-1][l-1]].s);else up(t+s[f[x-1][l-1]].s);
if(f[y+2][l-1]!=f[y+2][l])up(t+s[f[y+2][l]].s+s[f[x+1][l+1]].s);else up(t+s[f[x+1][l+1]].s);
}else{
if(f[y+1][l-2]!=f[y][l-2])up(t+s[f[y][l-2]].s);
if(f[y+2][l-1]!=f[y+2][l])up(t+s[f[y+2][l]].s);
}
}
if(y<n&&r<n&&!f[y+1][r+1]){
t=s[i].s+2+s[f[y+2][r+1]].s+s[f[y+1][r+2]].s;
if(x==y&&l==r){
if(f[y+1][r+2]!=f[y][r+2])up(t+s[f[y][r+2]].s+s[f[x-1][l+1]].s);else up(t+s[f[x-1][l+1]].s);
if(f[y+2][r+1]!=f[y+2][r])up(t+s[f[y+2][r]].s+s[f[x+1][l-1]].s);else up(t+s[f[x+1][l-1]].s);
}else{
if(f[y+1][r+2]!=f[y][r+2])up(t+s[f[y][r+2]].s);
if(f[y+2][r+1]!=f[y+2][r])up(t+s[f[y+2][r]].s);
}
}
}
return printf("%d",ans),0;
}