【洛谷 P6624 [省选联考 2020 A 卷] 作业题】【矩阵树定理】

题意

给一个$n$个点$m$条边的简单无向图，定义一棵生成树的权值为其边权和与边权$\gcd$的乘积。求所有生成树的权值和。
$n\le 30,w_i\le 152501$

分析

将$\gcd$拆成欧拉函数求和的形式，得到$$ans=\sum _w\phi (w)\ast [所有边权都是w倍数的生成树权值和]$$

问题转化为如何求所有生成树的权值和。

最暴力的方法是枚举一条边，强制这条边在生成树中，然后把基尔霍夫矩阵中这条边所在的行和列删掉，对剩下的$n-1$阶子式求行列式，再乘上该边的边权，求和就是答案。那么我们可以把一条边的边权看做多项式$1+w_{i}x$，然后求得行列式的一次项系数就是答案。因为若在行列式中选取$w_{i}x$这一项，等价于强制这条边在生成树中，并将生成树数量乘上边权加入到答案中。因此只需要在模$x^2$意义下求行列式即可。

对于时间复杂度，因为每条边最多被加入${\sigma }_0(w_i)$次，且若某次选出来的边数小于$n-1$，显然贡献必然为$0$。因此复杂度的一个上界为$O(n^5\frac{{\sigma }_0(w_i)}{n-1})=O(n^4{\sigma }_0(w_i))$。

代码

#include
using namespace std;

typedef long long LL;

const int N = 35;
const int M = 160005;
const int MOD = 998244353;

int n, m, tot, prime[M], phi[M], p1[N * N], p2[N * N];
bool not_prime[M];
vector<int> edg[M];

int ksm(int x, int y)
{
	int ans = 1;
	while (y)
	{
		if (y & 1) ans = (LL)ans * x % MOD;
		x = (LL)x * x % MOD; y >>= 1;
	}
	return ans;
}

struct poly
{
	int x, y;
	poly operator + (const poly & a) {return (poly){(x + a.x) % MOD, (y + a.y) % MOD};}
	poly operator - (const poly & a) {return (poly){(x - a.x) % MOD, (y - a.y) % MOD};}
	poly operator * (const poly & a) {return (poly){(LL)x * a.x % MOD, ((LL)x * a.y + (LL)y * a.x) % MOD};}
	poly operator * (const int & a) {return (poly){(LL)x * a % MOD, (LL)y * a % MOD};}
	poly inv() {int w = ksm(x, MOD - 2); return (poly){w, (LL)-w * w % MOD * y % MOD};}
}a[N][N];

void get_prime(int n)
{
	phi[1] = 1;
	for (int i = 2; i <= n; i++)
	{
		if (!not_prime[i]) prime[++tot] = i, phi[i] = i - 1;
		for (int j = 1; j <= tot && i * prime[j] <= n; j++)
		{
			not_prime[i * prime[j]] = 1;
			if (i % prime[j] == 0)
			{
				phi[i * prime[j]] = phi[i] * prime[j];
				break;
			}
			phi[i * prime[j]] = phi[i] * (prime[j] - 1);
		}
	}
}

poly solve(int n)
{
	poly ans = (poly){1, 0};
	for (int i = 1; i <= n; i++)
	{
		int pos = i;
		for (int j = i; j <= n; j++)
			if (a[j][i].x || a[j][i].y) {pos = j; break;}
		if (pos != i)
		{
			ans = ans * (-1);
			for (int j = i; j <= n; j++)
				swap(a[i][j], a[pos][j]);
		}
		ans = ans * a[i][i];
		for (int j = i + 1; j <= n; j++) if (a[j][i].x || a[j][i].y)
		{
			poly w = a[j][i] * a[i][i].inv();
			for (int k = i; k <= n; k++)
				a[j][k] = a[j][k] - a[i][k] * w;
		}
	}
	return ans;
}

int main()
{
	scanf("%d%d", &n, &m);
	int mx = 0;
	for (int i = 1; i <= m; i++)
	{
		int w; scanf("%d%d%d", &p1[i], &p2[i], &w);
		edg[w].push_back(i);
		mx = max(mx, w);
	}
	get_prime(mx);
	int ans = 0;
	for (int i = 1; i <= mx; i++)
	{
		int sum = 0;
		for (int j = i; j <= mx; j += i) sum += edg[j].size();
		if (sum < n - 1) continue;
		for (int j = 1; j <= n; j++)
			for (int k = 1; k <= n; k++)
				a[j][k] = (poly){0, 0};
		for (int j = i; j <= mx; j += i)
			for (int id : edg[j])
			{
				int x = p1[id], y = p2[id];
				poly t = (poly){1, j};
				a[x][x] = a[x][x] + t; a[y][y] = a[y][y] + t;
				a[x][y] = a[x][y] - t; a[y][x] = a[y][x] - t;
			}
		poly res = solve(n - 1);
		(ans += (LL)res.y * phi[i] % MOD) %= MOD;
	}
	printf("%d\n", (ans + MOD) % MOD);
	return 0;
}