The recursion is obvious ... But if you want to do matrix multiplication, you have to split the points .. At first, I was very confused about every weight v>1 All sides are new v-1 There are two nodes to transfer ... And then TLE 了 ... Break each point into 9 Just one ... Time complexity O((9N)^3*logT)

--------------------------------------------------------------------------

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn = 900;
const int MOD = 2009;
const int lim = 46300;

typedef int MATRIX[maxn][maxn];

MATRIX mat, Q, tmp;
int N, T, V, Id[maxn][10];
char s[maxn];

inline void upd(int &x, int t) {
if((x += t) >= lim) x %= MOD;
}

void MUL(MATRIX& A, MATRIX& B) {
memset(tmp, 0, sizeof tmp);
for(int i = 0; i < V; i++)
for(int k = 0; k < V; k++)
for(int j = 0; j < V; j++)
upd(tmp[i][j], A[i][k] * B[k][j]);
memcpy(A, tmp, sizeof A);
}

int main() {
scanf("%d%d", &N, &T);
V = 0;
for(int i = 0; i < N; i++)
for(int j = 0; j < 9; j++)
Id[i][j] = V++;
memset(mat, 0, sizeof mat);
for(int i = 0; i < N; i++) {
scanf("%s", s);
for(int j = 0; j < N; j++)
if(s[j] > '0') mat[Id[j][0]][Id[i][s[j] - '1']] = 1;
}
for(int i = 0; i < N; i++)
for(int j = 1; j < 9; j++)
mat[Id[i][j]][Id[i][j - 1]] = 1;
memset(Q, 0, sizeof Q);
for(int i = 0; i < V; i++) Q[i][i] = 1;
for(; T; T >>= 1, MUL(mat, mat))
if(T & 1) MUL(Q, mat);
printf("%d\n", Q[Id[N - 1][0]][0] % MOD);
return 0;
}

--------------------------------------------------------------------------

## 1297: [SCOI2009] Get lost

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 857  Solved: 602
[Submit][Status][Discuss]

## Description

windy Lost in digraph . The directed graph has N Nodes ,windy From the node 0 set out , He has to happen to be T Time to node N-1. Now let's give the digraph , You can tell windy How many different paths are there in all ？ Be careful ：windy Can't linger on a node , And the time passing through a directed edge is strictly a given time .

## Input

The first line contains two integers ,N T. Next there is N That's ok , One length per line is N String . The first i Xing di j As a '0' Represents slave node i To the node j There is no side . by '1' To '9' Represents slave node i To the node j It takes time .

## Output

Contains an integer , Number of possible paths , This number can be very large , Just output this number divided by 2009 The remainder of .

## Sample Input

【 Enter sample one 】
2 2
11
00

【 Enter example 2 】
5 30
12045
07105
47805
12024
12345

## Sample Output

【 Output sample one 】
1

【 Examples explain one 】
0->0->1

【 Output sample 2 】
852

## HINT

30% The data of , Satisfy 2 <= N <= 5 ; 1 <= T <= 30 .
100% The data of , Satisfy 2 <= N <= 10 ; 1 <= T <= 1000000000 .

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