[Gym-102020H]Hyperpath（矩阵快速幂）

标签：amp   离散   pre   end   连通图   ase   res   int   operator   

离散数学里面好像有一个连通图的可达矩阵，通过矩阵的幂乘找到该点每一步可到达的点
该题明显是k步后到了哪些点

int mod;
struct mat {
	int data[110][110] = { 0 };
	int n = 110;

	mat() {};
	mat(int n) : n(n) {};
	mat(int d, int x) {
		n = d;
		for (int i = 0; i < n; ++i)	data[i][i] = x;
	}
	 
	void disp(int n) {
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n; ++j)
				printf("%d%c", data[i][j], j == n - 1 ? ‘\n‘ : ‘ ‘);
	}

	int* operator [](int i) {
		return data[i];
	}

	mat operator * (mat& other) {
		mat res(n);
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < n; ++j) {
				for (int k = 0; k < n; ++k) {
					res[i][j] += 1ll *  data[i][k] * other[k][j] % mod;
					if (mod <= res[i][j])	res[i][j] -= mod;
				}
			}
		}
		return res;
	}
};


mat qpow(mat base, ll n) {
	mat res(base.n, 1);
	while (n) {
		if (n & 1)	res = res * base;
		base = base * base;
		n >>= 1;
	}
	return res;
}

signed main() {
	int n, m, st, ed;
	ll k;
	scanf("%d %d %d", &n, &m, &mod);
	scanf("%lld %d %d", &k, &st, &ed);

	mat g(n);
	for (int i = 0; i < m; ++i) {
		int u, v; scanf("%d %d", &u, &v);
		g[u][v] = 1;
		g[v][u] = 1;
	}

	mat ans = qpow(g, k);
	if (ans[st][ed] != 0)	printf("%d\n", ans[st][ed]);
	else {
		if (mod != 1e9 + 7)	mod = 1e9 + 7;
		else mod = 1e9 + 9;
		ans = qpow(g, k);
		puts(ans[st][ed] == 0 ? "Mendes will sleep in peace." : "0");
	}
	return 0;
}

[Gym-102020H]Hyperpath（矩阵快速幂）

标签：amp   离散   pre   end   连通图   ase   res   int   operator   

原文地址：https://www.cnblogs.com/wanshe-li/p/14492588.html