bzoj1297 [SCOI2009]迷路 矩阵快速幂

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=1297

题解

如果每一条边没有边权，那么就是一般的矩阵快速幂。

因为边权 (in [1, 9])，所以我们可以把每一个点拆成 (9) 个点，对于一条边权为 (w) 的边 (x o y)，可以建立一条边 (x_{w-1} o y_0)。

然后 (x_i o x_{i+1}) 连边。

然后矩阵快速幂就可以了。

---

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
	int f = 0, c;
	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
	x = c & 15;
	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
	f ? x = -x : 0;
}

const int N = 90 + 3;
const int P = 2009;

int n, m, T;
char s[N];

inline int smod(int x) { return x >= P ? x - P : x; }
inline void sadd(int &x, const int &y) { x += y; x >= P ? x -= P : x; }
inline int fpow(int x, int y) {
	int ans = 1;
	for (; y; y >>= 1, x = x * x % P) if (y & 1) ans = ans * x % P;
	return ans;
}

struct Matrix {
	int a[N][N];
	
	inline Matrix() { memset(a, 0, sizeof(a)); }
	inline Matrix(const int &x) {
		memset(a, 0, sizeof(a));
		for (int i = 1; i <= m; ++i) a[i][i] = x;
	}
	
	inline Matrix operator * (const Matrix &b) {
		Matrix c;
		for (int k = 1; k <= m; ++k)
			for (int i = 1; i <= m; ++i)
				for (int j = 1; j <= m; ++j)
					sadd(c.a[i][j], a[i][k] * b.a[k][j] % P);
		return c;
	}
} A;

inline Matrix fpow(Matrix x, int y) {
	Matrix ans(1);
	for (; y; y >>= 1, x = x * x) if (y & 1) ans = ans * x;
	return ans;
}

inline void work() {
	m = 9 * n;
	A = fpow(A, T);
	printf("%d
", A.a[1][n]);
}

inline void init() {
	read(n), read(T);
	for (int i = 1; i <= n; ++i) {
		scanf("%s", s + 1);
		for (int j = 1; j <= n; ++j)
			if (s[j] != '0') A.a[i + (s[j] - '1') * n][j] = 1;
	}
	for (int i = 1; i <= n; ++i)
		for (int j = 0; j < 9 - 1; ++j)
			A.a[i + j * n][i + (j + 1) * n] = 1;
}

int main() {
#ifdef hzhkk
	freopen("hkk.in", "r", stdin);
#endif
	init();
	work();
	fclose(stdin), fclose(stdout);
	return 0;
}