DPR Walk

题意

给定一个无向图，求路径长度为 \(k\) 的路径条数。

\(n \le 50\)。

Sol

考虑 \(dp\)，设 \(f_{i, j}\) 表示从 \(i \to j\) 的路径长为 \(k\) 的方案数。

不难发现转移即为矩阵乘法。

直接快速幂即可。

Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#define int long long
using namespace std;
#ifdef ONLINE_JUDGE

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;

#endif
int read() {
	int p = 0, flg = 1;
	char c = getchar();
	while (c < '0' || c > '9') {
		if (c == '-') flg = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9') {
		p = p * 10 + c - '0';
		c = getchar();
	}
	return p * flg;
}
void write(int x) {
	if (x < 0) {
		x = -x;
		putchar('-');
	}
	if (x > 9) {
		write(x / 10);
	}
	putchar(x % 10 + '0');
}
const int N = 55, mod = 1e9 + 7;

void Mod(int &x) {
	if (x >= mod) x -= mod;
	if (x < 0) x += mod;
}

struct Matrix {

int A[N][N];
int n, m;

int* operator [](int x) {
	return A[x];
}

Matrix (int n_, int m_) {
	for (int i = 1; i <= n_; i++)
		for (int j = 1; j <= m_; j++)
			A[i][j] = 0;
	n = n_, m = m_;
}

Matrix (int n_, int m_, int flg) {
	for (int i = 1; i <= n_; i++)
		for (int j = 1; j <= m_; j++)
			A[i][j] = 0;
	for (int i = 1; i <= n_; i++)
		A[i][i] = 1;
	n = n_, m = m_;
}

friend Matrix operator *(Matrix a, Matrix b) {
	Matrix ans(a.n, b.m);
	for (int i = 1; i <= a.n; i++)
		for (int j = 1; j <= a.m; j++)
			for (int k = 1; k <= b.m; k++)
				ans[i][j] += a[i][k] * b[k][j] % mod, Mod(ans[i][j]);
	return ans;
}

friend Matrix operator ^(Matrix x, int k) {
	Matrix ans(x.n, x.m, 1);
	while (k) {
		if (k & 1) ans = ans * x;
		x = x * x;
		k >>= 1;
	}
	return ans;
}

};

signed main() {
	int n = read(), k = read();
	Matrix T(n, n);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			T[i][j] = read();
	T = T ^ k;
	int ans = 0;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			ans += T[i][j], Mod(ans);
	write(ans), puts("");
	return 0;
}