bzoj1297 / P4159 [SCOI2009]迷路

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #define re register
 5 using namespace std;
 6 const int mod=2009;
 7 int m,n,t;long long q;
 8 struct matrix{
 9     int a[92][92];
10     matrix(){memset(a,0,sizeof(a));}
11     matrix operator * (const matrix &tmp) const{
12         matrix c;
13         for(int i=1;i<=n;++i)
14             for(int j=1;j<=n;++j)
15                 for(int k=1;k<=n;++k)
16                     c.a[i][j]=(c.a[i][j]+a[i][k]*tmp.a[k][j])%mod;
17         return c;
18     }
19     matrix Pow(matrix x,int y){
20         matrix res;
21         for(int i=1;i<=n;++i) res.a[i][i]=1;
22         for(;y;y>>=1,x=x*x)
23             if(y&1) res=res*x;
24         return res;
25     }
26 }st;
27 int idx(int x,int y){return x+m*y;}//新点编号
28 int main(){
29     scanf("%d%d",&m,&t);n=m*9;
30     for(int i=1;i<=m;++i){
31         for(int j=1;j<=8;++j)
32             st.a[idx(i,j)][idx(i,j-1)]=1;//规定idx(i,0)作为原图的点向其他点连边，其他点与该点的边权就转化为连到idx(i,dist-1)上
33         scanf("%lld",&q);
34         for(int j=m;j>=1;--j,q/=10)
35             if(q%10) st.a[i][idx(j,q%10-1)]=1;
36     }st=st.Pow(st,t);
37     printf("%d",st.a[1][m]);
38     return 0;
39 }