[CF917D]Stranger Trees[矩阵树定理+解线性方程组]

题意

给你 \(n\) 个点的无向完全图，指定一棵树 \(S\)，问有多少棵生成树和这棵树的公共边数量为 \(k\in[0,n-1]\)

\(n\leq 100\)

分析

• 考虑矩阵树定理，把对应的树边的边权设置成 \(x\) 然后构造基尔霍夫矩阵， 结果记为 \(val\) ，有

  \[val=\sum_\limits{i=0}^{n-1}x^ians_i
  \]

  其中 \(ans_i\) 表示和 \(S\) 的公共边数量为 \(i\) 的生成树的个数。

• 发现这是一个关于 \(x\) 的多项式，我们要求每一项的系数 \(ans_i\) ，所以搞出 \(x\in[0, n -1]\) 的 \(val\) 然后高斯消元即可。

• 总时间复杂度为 \(O(n^4)\)。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define go(u) for(int i = head[u], v = e[i].to; i; i=e[i].lst, v=e[i].to)
#define rep(i, a, b) for(int i = a; i <= b; ++i)
#define pb push_back
inline int gi() {
    int x = 0,f = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while(isdigit(ch)) {
        x = (x << 3) + (x << 1) + ch - 48;
        ch = getchar();
    }
    return x * f;
}
template <typename T> inline void Max(T &a, T b){if(a < b) a = b;}
template <typename T> inline void Min(T &a, T b){if(a > b) a = b;}
const int N = 104, mod = 1e9 + 7;
int n;
LL a[N][N], G[N][N], gg[N][N];
LL Pow(LL a, LL b) {
	LL res = 1ll;
	for(; b; b >>= 1, a = a * a % mod) if(b & 1) res = res * a % mod;
	return res;
}
void Gauss(int n, int m, LL (*G)[N]) {
	for(int u =0, col = 0; col <= m; ++col, ++u) {
		int sel = u;
		for(;sel <= n && !G[sel][col]; ++sel);
		if(sel > n) { --u; continue;}
		if(sel ^ u) {for(int i = 1; i <= m + 1; ++i) swap(G[u][i], G[sel][i]);}
		LL inv = Pow(G[u][col], mod - 2);
		for(int i = col; i <= m + 1; ++i) G[u][i] = G[u][i] * inv % mod;
		for(int v = 1; v <=n; ++v)if(u ^ v) {
			LL x = G[v][col];
			for(int i = col; i <= m + 1; ++i) G[v][i] = ((G[v][i] - G[u][i] * x) % mod + mod) % mod;
		}
	}
}
LL det(int n, int m, LL (*G)[N]) {
	LL c = 0;
	for(int u = 2, col = 2; col <= m; ++col, ++u) {
		int sel = u;
		for(;sel <= n && !G[sel][col]; ++sel);
		if(sel > n) { u--; continue;}
		if(sel ^ u) {c ^= 1; for(int i = 1; i <= m; ++i) swap(G[u][i], G[sel][i]);}

		for(int v = u + 1; v <= n; ++v)
		while(G[v][col]) {
			LL x = G[v][col] / G[u][col];
			for(int i = col; i <= m; ++i) G[v][i] = ((G[v][i] - x * G[u][i])%mod + mod) % mod;
			if(!G[v][col]) break;
			c ^= 1;
			for(int i = 1; i <= m; ++i) swap(G[u][i], G[v][i]);
		}
	}
	LL ans = 1ll;
	for(int i = 2; i <= n; ++i) ans = ans * G[i][i] % mod;
	if(c) ans = mod - ans;
	return ans;
}
int main() {
	n = gi();
	rep(i, 1, n - 1){
		int u = gi(), v = gi();
		a[u][v] ++, a[v][u] ++;
	}
	rep(x, 0, n - 1) {
		memset(G, 0, sizeof G);
		rep(i, 1, n)
		rep(j, 1, n) {
			G[j][j] += (a[i][j] ? x : 1);
			G[i][j] -= (a[i][j] ? x : 1);
		}
		rep(i, 1, n)
		rep(j, 1, n) if(G[i][j] < 0) G[i][j] += mod;
		gg[x][n] = det(n, n, G);
		LL tmp = 1;
		for(int i = 0; i < n; ++i, tmp = tmp * x % mod) gg[x][i] = tmp;
	}
	Gauss(n - 1, n - 1, gg);
	for(int i = 0; i < n; ++i) printf("%lld%c", gg[i][n], i == n?'\n':' ');
	return 0;
}