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(egin{bmatrix}cnt & a\0 & bend{bmatrix}),只需要求得(Ans=egin{bmatrix}cnt & a\0 & bend{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
");
return 0;
}
if (!h1&&!h2){
printf("%d
",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
",(Ans.v[0][0]-Ans.v[0][1]+p)%p);
return 0;
}