Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4159 [SCOI2009] 迷路

P4159 [SCOI2009] 迷路

考虑到如果边权为 $1$ ,那么显然直接矩阵加速即可。注意的边权很小,考虑拆点,每个点拆为若干个点,表示还有几秒到达,直接上矩阵加速即可。

查看代码
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#include <cstdio>
using namespace std;
const int N = 110, mod = 2e3 + 9;
struct Matrix
{
int n, m, w[N][N];
Matrix()
{
}
Matrix(int x, int y)
{
n = x, m = y;
}
Matrix(int x, int y, int k)
{
n = x, m = y;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
w[i][j] = i == j ? k : 0;
}
friend Matrix operator*(Matrix x, Matrix y)
{
Matrix res(x.n, y.m, 0);
for (int i = 1; i <= res.n; i++)
for (int j = 1; j <= res.m; j++)
for (int k = 1; k <= x.m; k++)
(res.w[i][j] += x.w[i][k] * y.w[k][j] % mod) %= mod;
return res;
}
friend Matrix operator^(Matrix b, int k)
{
Matrix res(b.n, b.m, 1);
while (k)
{
k & 1 && (res = res * b, 0);
b = b * b;
k >>= 1;
}
return res;
}
};
int num (int a, int b)
{
return (a - 1) * 9 + b;
}
int main ()
{
int n, m;
scanf("%d%d", &n, &m);
static Matrix mat(n * 9, n * 9);
for (int i = 1; i <= n; i++)
for (int j = 1, a; j <= n; j++)
{
scanf("%1d", &a);
a && (mat.w[num(i, 1)][num(j, a)]++);
}
for (int i = 1; i <= n ;i++)
for (int j = 2; j <= 9; j++)
mat.w[num(i, j)][num(i, j - 1)]++;
printf("%d", (mat ^ m).w[num(1, 1)][num(n, 1)]);
return 0;
}