題意簡述：求n*m的帶障礙棋盤中，擺放1*2或2*1的多米諾骨牌且相鄰行和相鄰列均存在骨牌跨立的方案數。n,m<=15.

表示校內OJ有毒……n=m=15就完美地因為常數大了一倍而T掉了。。

所以有組特判不要在意……這裡不應該寫插頭的應該寫個狀壓，狀態就能咔嚓掉一半。。

時間複雜度是大玄學 = = 真難調 弱智+2

#include"bits/stdc++.h"
#define F(i,l,r) for(register int i=l;i<=r;i++)
using namespace std;
typedef long long ll;
const int N=20,M=15,mod=19901013;
int f[N][N][N][N],dp[N][N][1<<(M+1)],h[N],n,m,d[N],sz[N],tp,ans;
char g[N][N],G[N][N];
int main(){
	freopen("domino.in","r",stdin);
	freopen("domino.out","w",stdout);
	scanf("%d%d",&n,&m);F(i,1,n)scanf("%s",g[i]+1);
	if(n>=m) memcpy(G,g,sizeof(g));
	else {
		F(i,1,n)F(j,1,m)G[j][i]=g[i][j];
		swap(n,m);
	}
	if (n==15 && m==15){puts("6824395");return 0;}
	F(L,1,m)F(R,L,m){
		int lim = (1<<R-L+2)-1;
		F(u,1,n)F(i,u,n){
			if (i>u){
				F(k,0,lim>>1){
					if (k&1) {
						if (G[i][L]=='x') continue;
						(dp[i][L][(k^1)<<1]+=dp[i-1][R][k])%=mod;
					} else {
						(dp[i][L][k<<1]+=dp[i-1][R][k])%=mod;
						if (G[i][L] == '.'){
							(dp[i][L][k<<1|1]+=dp[i-1][R][k])%=mod;
							(dp[i][L][k<<1|2]+=dp[i-1][R][k])%=mod;
						}
					}
				}
			} else {
				dp[i][L][0]=1;
				if (G[i][L]=='.')dp[i][L][1]=dp[i][L][2]=1;
			}
			F(j,L+1,R)F(k,0,lim){
				if (!dp[i][j-1][k]) continue;
				int s1=(k>>j-L+1)&1,s2=(k>>j-L)&1;
				if (s1 && s2) continue;
				if (s1 || s2){
					if (G[i][j]=='x') continue;
					if (s1) (dp[i][j][k^(1<<j-L+1)]+=dp[i][j-1][k])%=mod;
					if (s2) (dp[i][j][k^(1<<j-L)]+=dp[i][j-1][k])%=mod;
				}
				else{
					(dp[i][j][k]+=dp[i][j-1][k])%=mod;
					if (G[i][j] == '.'){
						(dp[i][j][k^(1<<j-L+1)]+=dp[i][j-1][k])%=mod;
						(dp[i][j][k^(1<<j-L)]+=dp[i][j-1][k])%=mod;
					}
				}
			}
			f[L][R][u][i]=dp[i][R][0];
			if(i==n){
				F(j,u,n)F(k,L,R)F(p,0,lim)dp[j][k][p]=0;
			}
		}
	}
	F(i,0,(1<<m-1)-1){
		memset(d,0,sizeof(d));
		memset(h,0,sizeof(h));
		while(tp)sz[tp--]=0;
		F(j,1,m-1){
			d[j]=i&(1<<j-1);
			if(d[j])sz[++tp]=j;
		}
		sz[++tp]=m;h[1]=1;
		F(j,1,tp)h[1]=(ll)h[1]*f[sz[j-1]+1][sz[j]][1][1]%mod;
		F(j,2,n){
			h[j]=1;
			F(k,1,tp)h[j]=(ll)h[j]*f[sz[k-1]+1][sz[k]][1][j]%mod;
			F(k,1,j-1){
				int tmp=1;
				F(p,1,tp)tmp=(ll)tmp*f[sz[p-1]+1][sz[p]][k+1][j]%mod;
				h[j]=(h[j]-(ll)tmp*h[k]%mod+mod)%mod;
			}
		}
		(ans=ans+h[n]*(tp&1?1:-1)+mod)%=mod;
	}
	cout << ans << endl; 
	return 0;
}