bzoj 1297: [SCOI2009]迷路
思路：首先我们先了解一个性质， 对于一个邻接矩阵 edge 来说， (edge^t)[ i ][ j ] 表示走t次能从i 到 j 的路径条数，但是
这道题里面不是邻接矩阵，所以我们要拆点变成邻接矩阵然后矩阵快速幂。
1 #include<bits/stdc++.h> 2 #define LL long long 3 #define ll long long 4 #define fi first 5 #define se second 6 #define mk make_pair 7 #define pii pair<int,int> 8 #define piii pair<int, pair<int,int>> 9 10 using namespace std; 11 12 const int N = 100 + 7; 13 const int inf = 0x3f3f3f3f; 14 const LL INF = 0x3f3f3f3f3f3f3f3f; 15 const int mod = 1e9 + 7; 16 const int Mod = 2009; 17 18 struct Matrix 19 { 20 int r, c; 21 int a[N][N]; 22 Matrix(int r = 0, int c = 0) 23 { 24 this -> r = r; 25 this -> c = c; 26 memset(a,0,sizeof(a)); 27 } 28 void init() 29 { 30 for(int i=0;i<r;i++) 31 for(int j=0;j<c;j++) 32 a[i][j]=(i==j); 33 } 34 Matrix operator * (const Matrix &B)const 35 { 36 Matrix C(r, c); 37 for(int i=0;i<r;i++) 38 for(int j=0;j<c;j++) 39 for(int k=0;k<r;k++) 40 C.a[i][j]=(C.a[i][j]+1LL*a[i][k]*B.a[k][j])%Mod; 41 return C; 42 } 43 Matrix operator ^ (const ll &p)const 44 { 45 Matrix A=(*this),res(r, c); 46 res.init(); 47 ll t=p; 48 while(t) 49 { 50 if(t&1)res=res*A; 51 A=A*A; 52 t>>=1; 53 } 54 return res; 55 } 56 }M; 57 58 59 int n, t; 60 int main() { 61 scanf("%d%d", &n, &t); 62 M.r = M.c = 9 * n; 63 for(int i = 0; i < n * 9; i++) 64 if(i % 9 != 0) M.a[i - 1][i] = 1; 65 66 for(int i = 0; i < n; i++) { 67 for(int j = 0; j < n; j++) { 68 int x; scanf("%1d", &x); 69 if(!x) continue; 70 M.a[i * 9 + x - 1][j * 9] = 1; 71 } 72 } 73 74 M = M ^ t; 75 76 printf("%d\n", M.a[0][(n - 1) * 9]); 77 return 0; 78 } 79 80 /* 81 */
