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  • loj 6391 「THUPC2018」淘米神的树 / Tommy

    题目传送门

      传送门

      当 $a = b$ 的时候答案是:

    $$
    n! prod_i frac{1}{sz_i}
    $$

      现在考虑 $a eq b$ 的情形,考虑 $a ightarrow b$ 的链。不妨设这些点依次是 $v_1 = a, v_1, cdots, v_m = b$。

      注意到硬点 $v_k$ 是链上最后一个点被染红的点非常难算。

      考虑忽略掉其中某条边,然后硬点它右侧的点满足 $v_{i + 1}$ 在 $v_i$ 之前被染红,左侧的点满足 $v_i$ 在 $v_{i + 1} $ 之前被染红。如果认为 $v_0$ 和 $v_{1}$ 有边相连, $v_{m}$ 和 $v_{m + 1}$ 有边相连。这些边每条边被忽略的方案数之和等于所求的方案数的两倍。

      考虑怎么求这些方案数。注意到断掉一条边只会改变这条链上的点的 size。设断掉链上所有边后,前 $i$ 个点的 size 之和为 $s_i$,那么断掉 $v_i$ 和 $v_{i + 1}$ 之间的边,相当于是求 $prod_{j = 0, i eq j}^m |s_i - s_j|$。讨论一下拆掉绝对值,然后用洛必达法则多点求值就行了。

      时间复杂度 $O(nlog^2 n)$

    Code

    /**
     * loj
     * Problem#6391
     * Accepted
     * Time: 27112ms
     * Memory: 80392k
     */
    #include <bits/stdc++.h>
    using namespace std;
    typedef bool boolean;
    
    #define ll long long
    
    template <typename T>
    void pfill(T* pst, const T* ped, T val) {
    	for ( ; pst != ped; *(pst++) = val);
    }
    
    template <typename T>
    void pcopy(T* pst, const T* ped, T* pval) {
    	for ( ; pst != ped; *(pst++) = *(pval++));
    }
    
    const int N = 524288;
    const int Mod = 998244353;
    const int bzmax = 20;
    const int g = 3;
    
    void exgcd(int a, int b, int& x, int& y) {
    	if (!b) {
    		x = 1, y = 0;
    	} else {
    		exgcd(b, a % b, y, x);
    		y -= (a / b) * x;
    	}
    }
    
    int inv(int a, int Mod) {
    	int x, y;
    	exgcd(a, Mod, x, y);
    	return (x < 0) ? (x + Mod) : (x);
    }
    
    template <const int Mod = :: Mod>
    class Z {
    	public:
    		int v;
    
    		Z() : v(0) {	}
    		Z(int x) : v(x){	}
    		Z(ll x) : v(x % Mod) {	}
    
    		friend Z operator + (const Z& a, const Z& b) {
    			int x;
    			return Z(((x = a.v + b.v) >= Mod) ? (x - Mod) : (x));
    		}
    		friend Z operator - (const Z& a, const Z& b) {
    			int x;
    			return Z(((x = a.v - b.v) < 0) ? (x + Mod) : (x));
    		}
    		friend Z operator * (const Z& a, const Z& b) {
    			return Z(a.v * 1ll * b.v);
    		}
    		friend Z operator ~ (const Z& a) {
    			return inv(a.v, Mod);
    		}
    		friend Z operator - (const Z& a) {
    			return Z(0) - a;
    		}
    		Z& operator += (Z b) {
    			return *this = *this + b;
    		}
    		Z& operator -= (Z b) {
    			return *this = *this - b;
    		}
    		Z& operator *= (Z b) {
    			return *this = *this * b;
    		}
    		friend boolean operator == (const Z& a, const Z& b) {
    			return a.v == b.v;
    		} 
    };
    
    typedef Z<> Zi;
    
    Zi qpow(Zi a, int p) {
    	if (p < Mod - 1)
    		p += Mod - 1;
    	Zi rt = 1, pa = a;
    	for ( ; p; p >>= 1, pa = pa * pa) {
    		if (p & 1) {
    			rt = rt * pa;
    		}
    	}
    	return rt;
    }
    
    const Zi inv2 ((Mod + 1) >> 1);
    
    class NTT {
    	private:
    		Zi gn[bzmax + 4], _gn[bzmax + 4];
    	public:
    		
    		NTT() {
    			for (int i = 0; i <= bzmax; i++) {
    				gn[i] = qpow(Zi(g), (Mod - 1) >> i);
    				_gn[i] = qpow(Zi(g), -((Mod - 1) >> i));
    			}
    		}
    
    		void operator () (Zi* f, int len, int sgn) {
    			for (int i = 1, j = len >> 1, k; i < len - 1; i++, j += k) {
    				if (i < j)
    					swap(f[i], f[j]);
    				for (k = len >> 1; k <= j; j -= k, k >>= 1);
    			}
    			
    			Zi *wn = (sgn > 0) ? (gn + 1) : (_gn + 1), w, a, b;
    			for (int l = 2, hl; l <= len; l <<= 1, wn++) {
    				hl = l >> 1, w = 1;
    				for (int i = 0; i < len; i += l, w = 1) {
    					for (int j = 0; j < hl; j++, w *= *wn) {
    						a = f[i + j], b = f[i + j + hl] * w;
    						f[i + j] = a + b;
    						f[i + j + hl] = a - b;
    					}
    				}
    			}
    
    			if (sgn < 0) {
    				Zi invlen = ~Zi(len);
    				for (int i = 0; i < len; i++) {
    					f[i] *= invlen;
    				}
    			}
    		}
    
    		int correct_len(int len) {
    			int m = 1;
    			for ( ; m <= len; m <<= 1);
    			return m;
    		}
    } NTT;
    
    void pol_inverse(Zi* f, Zi* g, int n) {
    	static Zi A[N];
    	if (n == 1) {
    		g[0] = ~f[0];
    	} else {
    		int hn = (n + 1) >> 1, t = NTT.correct_len(n << 1 | 1);
    		pol_inverse(f, g, hn);
    		
    		pcopy(A, A + n, f);
    		pfill(A + n, A + t, Zi(0));
    		pfill(g + hn, g + t, Zi(0));
    		NTT(A, t, 1);
    		NTT(g, t, 1);
    		for (int i = 0; i < t; i++) {
    			g[i] = g[i] * (Zi(2) - g[i] * A[i]);
    		}
    		NTT(g, t, -1);
    		pfill(g + n, g + t, Zi(0));
    	}
    }
    
    void pol_sqrt(Zi* f, Zi* g, int n) {
    	static Zi A[N], B[N];
    	if (n == 1) {
    		g[0] = f[0];
    	} else {
    		int hn = (n + 1) >> 1, t = NTT.correct_len(n + n);
    		
    		pol_sqrt(f, g, hn);
    
    		pfill(g + hn, g + n, Zi(0));
    		for (int i = 0; i < hn; i++)
    			A[i] = g[i] + g[i];
    		pfill(A + hn, A + t, Zi(0));
    		pol_inverse(A, B, n);
    		pcopy(A, A + n, f);
    		pfill(A + n, A + t, Zi(0));
    		NTT(A, t, 1);
    		NTT(B, t, 1);
    		for (int i = 0; i < t; i++)
    			A[i] *= B[i];
    		NTT(A, t, -1);
    		for (int i = 0; i < n; i++)
    			g[i] = g[i] * inv2 + A[i];
    	}
    }
    
    typedef class Poly : public vector<Zi> {
    	public:
    		using vector<Zi>::vector;
    
    		Poly& fix(int sz) {
    			resize(sz);
    			return *this;
    		}
    } Poly;
    
    Poly operator + (Poly A, Poly B) {
    	int n = A.size(), m = B.size();
    	int t = max(n, m);
    	A.resize(t), B.resize(t);
    	for (int i = 0; i < t; i++) {
    		A[i] += B[i];
    	}
    	return A;
    }
    
    Poly operator - (Poly A, Poly B) {
    	int n = A.size(), m = B.size();
    	int t = max(n, m);
    	A.resize(t), B.resize(t);
    	for (int i = 0; i < t; i++) {
    		A[i] -= B[i];
    	}
    	return A;
    }
    
    Poly sqrt(Poly a) {
    	Poly rt (a.size());
    	pol_sqrt(a.data(), rt.data(), a.size());
    	return rt;
    }
    
    Poly operator * (Poly A, Poly B) {
    	int n = A.size(), m = B.size();
    	int k = NTT.correct_len(n + m - 1);
    	if (n < 20 || m < 20) {
    		Poly rt (n + m - 1);
    		for (int i = 0; i < n; i++) {
    			for (int j = 0; j < m; j++) {
    				rt[i + j] += A[i] * B[j];
    			}
    		}
    		return rt;
    	}
    	A.resize(k), B.resize(k);
    	NTT(A.data(), k, 1);
    	NTT(B.data(), k, 1);
    	for (int i = 0; i < k; i++) {
    		A[i] *= B[i];
    	}
    	NTT(A.data(), k, -1);
    	A.resize(n + m - 1);
    	return A;
    }
    
    Poly operator ~ (Poly f) {
    	int n = f.size(), t = NTT.correct_len((n << 1) | 1);
    	Poly rt (t);
    	f.resize(t);
    	pol_inverse(f.data(), rt.data(), n);
    	rt.resize(n);
    	return rt;
    }
    
    Poly operator / (Poly A, Poly B) {
    	int n = A.size(), m = B.size();
    	if (n < m) {
    		return Poly {0};
    	}
    	int r = n - m + 1;
    	reverse(A.begin(), A.end());
    	reverse(B.begin(), B.end());
    	A.resize(r), B.resize(r);
    	A = A * ~B;
    	A.resize(r);
    	reverse(A.begin(), A.end());
    	return A;
    }
    
    Poly operator % (Poly A, Poly B) {
    	int n = A.size(), m = B.size();
    	if (n < m) {
    		return A;
    	}
    	if (m == 1) {
    		return Poly {0};
    	}
    	A = A - A / B * B;
    	A.resize(m - 1);
    	return A;
    }
    
    Zi Inv[N];
    void init_inv(int n) {
    	Inv[0] = 0, Inv[1] = 1;
    	for (int i = 2; i <= n; i++) {
    		Inv[i] = Inv[Mod % i] * Zi((Mod - (Mod / i)));
    	}
    }
    
    void diff(Poly& f) {
    	if (f.size() == 1) {
    		f[0] = 0;
    		return;
    	}
    	for (int i = 1; i < (signed) f.size(); i++) {
    		f[i - 1] = f[i] * Zi(i);
    	}
    	f.resize(f.size() - 1);
    }
    void integ(Poly& f) {
    	f.resize(f.size() + 1);
    	for (int i = (signed) f.size() - 1; i; i--) {
    		f[i] = f[i - 1] * Inv[i];
    	}
    	f[0] = 0;
    }
    
    Poly ln(Poly f) {
    	int n = f.size();
    	Poly h = f;
    	diff(h);
    	f = h * ~f;
    	f.resize(n - 1);
    	integ(f);
    	return f;
    }
    
    void pol_exp(Poly& f, Poly& g, int n) {
    	Poly h;
    	if (n == 1) {
    		g.resize(1);
    		g[0] = 1;
    	} else {
    		int hn = (n + 1) >> 1;
    		pol_exp(f, g, hn);
    		
    		h.resize(n), g.resize(n);
    		pcopy(h.data(), h.data() + n, f.data());
    
    		g = g * (Poly{1} - ln(g) + h);
    		g.resize(n);
    	}
    }
    
    Poly exp(Poly f) {
    	int n = f.size();
    	Poly rt;
    	pol_exp(f, rt, n);
    	return rt;
    }
    
    class PolyBuilder {
    	protected:
    		int num;
    		Poly P[N << 1];
    		
    		void _init(int *x, int l, int r) {
    			if (l == r) {
    				P[num++] = Poly{-Zi(x[l]), Zi(1)};
    				return;
    			}
    			int mid = (l + r) >> 1;
    			int curid = num++;
    			_init(x, l, mid);
    			int rid = num;
    			_init(x, mid + 1, r);
    			P[curid] = P[curid + 1] * P[rid];
    		}
    
    		void _evalute(Poly f, Zi* y, int l, int r) {
    			f = f % P[num++];
    			if (l == r) {
    				y[l] = f[0];
    				return;
    			}
    			int mid = (l + r) >> 1;
    			_evalute(f, y, l, mid);
    			_evalute(f, y, mid + 1, r);
    		}
    	public:
    		Poly evalute(Poly f, int* x, int n) {
    			Poly rt(n);
    			num = 0;
    			_init(x, 0, n - 1);
    			num = 0;
    			_evalute(f, rt.data(), 0, n - 1);
    			return rt;
    		}
    } PolyBuilder;
    
    Poly dividing(int *a, int l, int r) {
    	if (l == r)
    		return Poly {-Zi(a[l]), 1};
    	int mid = (l + r) >> 1;
    	return dividing(a, l, mid) * dividing(a, mid + 1, r);
    }
    
    int n, a, b;
    boolean on[N];
    int fa[N], sz[N];
    vector<int> G[N];
    
    void dfs(int p, int fa) {
    	sz[p] = 1;
    	::fa[p] = fa;
    	for (auto e : G[p]) {
    		if (e ^ fa) {
    			dfs(e, p);
    			sz[p] += sz[e];
    		}
    	}
    }
    
    int main() {
    	scanf("%d%d%d", &n, &a, &b);
    	for (int i = 1, u, v; i < n; i++) {
    		scanf("%d%d", &u, &v);
    		G[u].push_back(v);
    		G[v].push_back(u);
    	}
    	dfs(a, 0);
    	init_inv(n);
    	if (a == b) {
    		Zi ans = 1;
    		for (int i = 1; i <= n; i++)
    			ans = ans * i;
    		for (int i = 1; i <= n; i++)
    			ans *= Inv[sz[i]];
    		printf("%d
    ", ans.v);
    		return 0;
    	}
    	vector<int> c {0};
    	for (int p = b; p; p = fa[p]) {
    		on[p] = true;
    		c.push_back(sz[p]);
    	}
    	Poly F = dividing(c.data(), 0, (signed) c.size() - 1);
    	diff(F);
    	Poly y = PolyBuilder.evalute(F, c.data(), c.size());
    	Zi sgn = (F.size() & 1) ? (Zi(Mod - 1)) : Zi(1), sum = 0;
    	for (auto t : y) {
    		sum = sum + ~t * (sgn = -sgn);
    	}
    	for (int i = 1; i <= n; i++)
    		sum = sum * i;
    	for (int i = 1; i <= n; i++)
    		sum = sum * ((!on[i]) ? (Inv[sz[i]]) : Zi(1));
    	sum = sum * ((Mod + 1) >> 1);
    	printf("%d
    ", sum.v);
    	return 0;
    } 
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  • 原文地址:https://www.cnblogs.com/yyf0309/p/12346496.html
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