AtCoder Grand Contest 003 F - Fraction of Fractal

题目传送门：https://agc003.contest.atcoder.jp/tasks/agc003_f

题目大意：

给定一个\(H×W\)的黑白网格，保证黑格四连通且至少有一个黑格

定义分形如下：\(0\)级分形是一个\(1×1\)的黑色单元格，\(k+1\)级分形由\(k\)级分形得来。具体而言，\(k\)级分形中每个黑色单元格将会被替换为初始给定的\(H×W\)的黑白网格，每个白色单元格会被替换为\(H×W\)的全白网格

求\(k\)级分形的四连通分量数，答案对\(10^9+7\)取模

---

如果这个图上下联通且左右联通，那么答案即为1；如果上下左右都不联通，答案即为\(cnt^{k-1}\)，\(cnt\)为黑格个数

剩下的即为上下联通或左右联通，我们把它统一改为左右联通（你转一下就好了），统计\(cnt\)(黑格个数)，\(a\)(同一行相邻的黑块个数)，\(b\)(行联通个数)，于是我们构造矩阵\(\begin{bmatrix}cnt & a\\0 & b\end{bmatrix}\)，只需要求得​\(Ans=\begin{bmatrix}cnt & a\\0 & b\end{bmatrix}^{k-1}\)，答案即为\(Ans.v[1][1]-Ans.v[1][2]\)

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef pair<int,int> pii;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e3,p=1e9+7;
struct Matrix{
	int v[2][2];
	Matrix(){memset(v,0,sizeof(v));}
	void init(){for (int i=0;i<2;i++)	v[i][i]=1;}
}Ans;
Matrix operator *(const Matrix &x,const Matrix &y){
	Matrix z;
	for (int i=0;i<2;i++)
		for (int j=0;j<2;j++)
			for (int k=0;k<2;k++)
				z.v[i][k]=(z.v[i][k]+1ll*x.v[i][j]*y.v[j][k])%p;
	return z;
}
Matrix mlt(Matrix a,ll b){
	Matrix res; res.init();
	for (;b;b>>=1,a=a*a)	if (b&1)	res=res*a;
	return res;
}
int mlt(int a,ll b){
	int res=1;
	for (;b;b>>=1,a=1ll*a*a%p)	if (b&1)	res=1ll*res*a%p;
	return res;
}
char s[N+10][N+10];
int main(){
	int n=read(),m=read(); ll k; scanf("%lld",&k);
	for (int i=1;i<=n;i++)	scanf("%s",s[i]+1);
	int cnt=0,s1=0,s2=0,h1=0,h2=0;
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			if (s[i][j]=='#'){
				cnt++;
				if (s[i][j+1]=='#')	s1++;
				if (s[i+1][j]=='#')	s2++;
			}
		}
		if (s[i][1]=='#'&&s[i][m]=='#')	h1++;
	}
	for (int i=1;i<=m;i++)	if (s[1][i]=='#'&&s[n][i]=='#')	h2++;
	if (h1&&h2){
		printf("1\n");
		return 0;
	}
	if (!h1&&!h2){
		printf("%d\n",mlt(cnt,k-1));
		return 0;
	}
	if (!h1)	swap(s1,s2),swap(h1,h2);
	Ans.v[0][0]=cnt,Ans.v[0][1]=s1,Ans.v[1][1]=h1;
	Ans=mlt(Ans,k-1);
	printf("%d\n",(Ans.v[0][0]-Ans.v[0][1]+p)%p);
	return 0;
}