BZOJ 1297 迷路

因为边权最大为9，记录前9个状态矩阵快速幂。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxv 15
#define maxn 150
#define mod 2009
#define inf 100
using namespace std;
struct matrix
{
    int a[maxn][maxn];
}a,b;
int n,t,map[maxv][maxv],x,y,dp[maxv][10];
char s[maxv];
void pre_dp()
{
    dp[1][0]=1;
    for (int i=1;i<=9;i++)
    {
        for (int j=1;j<=n;j++)
            for (int k=1;k<=n;k++)
                if (map[k][j]<=i) dp[j][i]=(dp[j][i]+dp[k][i-map[k][j]])%mod;
    }
}
void get_table()
{
    int cnt=0;
    for (int i=1;i<=9;i++)
        for (int j=1;j<=n;j++)
            a.a[1][++cnt]=dp[j][i];
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
            if (map[i][j]!=inf)
                b.a[(9-map[i][j])*n+i][8*n+j]=1;
    int p1=n+1,p2=1;
    while (p1<=9*n)
    {
        b.a[p1][p2]=1;
        p1++;p2++;
    }
}
matrix mul(matrix a,matrix b)
{
    matrix c;
    for (int i=1;i<=9*n;i++)
        for (int j=1;j<=9*n;j++)
            c.a[i][j]=0;
    for (int i=1;i<=9*n;i++)
        for (int j=1;j<=9*n;j++)
            for (int k=1;k<=9*n;k++)
                c.a[i][j]=(c.a[i][j]+(a.a[i][k]*b.a[k][j])%mod)%mod;
    return c;
}
void f_pow(int y)
{
    while (y)
    {
        if (y&1) a=mul(a,b);
        b=mul(b,b);
        y>>=1;
    }
}
int main()
{
    scanf("%d%d",&n,&t);
    for (int i=1;i<=n;i++)
    {
        scanf("%s",s);
        for (int j=1;j<=n;j++)
        {
            map[i][j]=s[j-1]-'0';
            if (!map[i][j]) map[i][j]=inf;
        }
    }
    pre_dp();
    if (t<=9) {printf("%d\n",dp[n][t]%mod);return 0;}
    get_table();
    f_pow(t-9);
    printf("%d\n",a.a[1][9*n]%mod);
    return 0;
}