「CometOJ」Contest #11

Link

Aeon

显然字典序最大就是把最小的字母放在最后

Business

[动态规划]

简单dp

dp[i][j]dp[i][j]dp[i][j]表示到第iii天，当前有jjj块钱，最后返还的钱最多为多少

完全背包转移

Celebration

Description

有一个环

，求把它分成三段，使得每一段内无重复元素，且三段长度可以作为某个三角形的三边的方案数。

一个拆分方案可以看作一个三元组 (a,b,c)(a,b,c)(a,b,c)，其中 0<a<b<c≤n0lt alt b lt c le n0<a<b<c≤n，表示在第 a,b,ca,b,ca,b,c个位置之前断开。两个拆分不同当且仅当其对应的三元组不同。

n≤2×106nle 2times10^6n≤2×106

Solution

[计数] [树状数组]

定义长度不超过 n−12frac{n-1}{2}2n−1 ，且不含重复颜色的段为合法的段。记 prexpre_xprex 为以

为右端点的合法段最远的左端点，nxtxnxt_xnxtx 为以 xxx 为左端点的合法段最远的右端点。

先枚举题目中的aaa，那么b∈(a,nxta+1]bin(a, nxt_a + 1]b∈(a,nxta+1]。在确定了a,ba, ba,b的位置后，合法的ccc位于(b,nxtb+1](b, nxt_b + 1](b,nxtb+1]和[prea,n][pre_a, n][prea,n]的交集中

注意，这里的preapre_aprea必须是大于aaa的，即绕了一圈绕到右边去。否则一定不合法

可以用树状数组维护：

从左往右枚举aaa，把合法的bbb对应的(b,nxtb+1](b, nxt_b + 1](b,nxtb+1]这段区间在树状数组中+1；查询就直接查[prea,n][pre_a, n][prea,n]的区间和；在离开aaa的时候把(a,nxta+1](a, nxt_a + 1](a,nxta+1]区间-1

Code


#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back
#define DEBUG(x) cout << #x << " = " << x << endl;

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int  (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 2e6 + 10;

int N;
int A[Maxn];
int vis[Maxn];
int L[Maxn], R[Maxn];

inline int fix (int x) { return ((x - 1) % N + N) % N + 1; }

namespace BIT
{
	struct bit
	{
		LL sum[Maxn];
		inline void add (int x, int val) { for (; x <= N; x += x & (-x)) sum[x] += val; }
		inline LL query (int x) { LL ans = 0; for (; x; x -= x & (-x)) ans += sum[x]; return ans; } 
	} A, B;

	inline void update (int x, int y, int val)
	{
		if (x > y) return ;
		A.add (x, val * x), A.add (y + 1, -val * (y + 1));
		B.add (x, val), B.add (y + 1, -val);
	}

	inline LL query (int x) { return B.query (x) * (x + 1) - A.query (x); }

	inline LL query (int x, int y) { if (x > y) return 0; return query (y) - query (x - 1); }
}

inline void Init ()
{
	int r = 0;
	for (int i = 1; i <= N; ++i)
	{
		while (r < N && !vis[A[r + 1]]) ++r, ++vis[A[r]];
		R[i] = min (r, i + (N - 1) / 2 - 1);
		if (vis[A[i]]) --vis[A[i]];
	}

	memset (vis, 0, sizeof vis);
	int l = 1;
	while (!vis[A[fix (l - 1)]]) l = fix (l - 1), ++vis[A[l]];
	L[1] = fix (max (l, N - (N - 1) / 2 + 1));

	for (int i = 2; i <= (N - 1) / 2; ++i)
	{
		while (vis[A[i - 1]]) --vis[A[l]], l = fix (l + 1);
		if (l < i) break;
		L[i] = max (l, fix ((i - 1 + N) - (N - 1) / 2 + 1));
		++vis[A[i - 1]];
	}
}

inline void Solve ()
{
	Init ();

	LL ans = 0;
	int	p = 1;
	for (int i = 1; i <= N; ++i)
	{
		if (L[i] < i) break;
		while (p < N && p + 1 <= R[i] + 1)
		{
			++p;
			BIT :: update (p + 1, R[p] + 1, 1);
		}
		ans += BIT :: query (L[i], N);
		BIT :: update ((i + 1) + 1, R[i + 1] + 1, -1);
	}

	cout << ans << endl;
}

inline void Input ()
{
	N = read<int>();
	for (int i = 1; i <= N; ++i) A[i] = read<int>();
}

int main()
{

#ifdef hk_cnyali
	freopen("C.in", "r", stdin);
	freopen("C.out", "w", stdout);
#endif

	Input ();
	Solve ();

	return 0;
}

Disaster

[kruskal重构树]

kruskal重构树模板题

Effort

Description

有 mmm种数据结构(可把数据结构想像成游戏中的种族)，第 iii种有 aia_iai个，每个可给敌人造成至少 111 次，至多 bib_ibi次伤害。有n nn名敌人，每人承担至少一次伤害。求总情况数模 998244353998244353998244353的值

数据结构两两不同 (同种的任两个也不同)，敌人两两不同。

两种方案不同当且仅当某个数据结构造成的伤害不同，或某个敌人受到的伤害不同。

n×m≤105,ai≤105,bi <998244353n times mle 10^5, a_ile 10^5, b_i < 998244353n×m≤105,ai≤105,bi <998244353

Solution

[组合数学] [生成函数] [多项式] [NTT]

注意到每个敌人的伤害是无序的，即这个敌人在被第几次攻击到都是一样的，而其他限制都是有序的

于是可以钦定攻击顺序和受到伤害的顺序。形象地理解就是先把所有数据结构按顺序摆一排，确定每个数据结构攻击多少次，这样就确定出一个攻击序列。再在这个攻击序列上插n−1n-1n−1个板就是这个攻击序列方案数

看上去这是由两个部分构成的（先确定攻击序列，再插板），但实际上可以同时算

设Fi(x)F_i(x)Fi(x)表示一个第iii种数据结构插板方案的生成函数，它的kkk次项系数表示插kkk个板的方案数。那么[xk]Fiai(x)displaystyle [x^k]F_i^{a_i}(x)[xk]Fiai(x)就是在第iii种里插kkk个板的方案数了

考虑如何求Fi(x)F_i(x)Fi(x)

第kkk项系数其实就是

(1k)+(2k)+⋯+(bik)
binom{1}{k} + binom{2}{k} + cdots + binom{b_i}{k}
(k1)+(k2)+⋯+(kbi)

枚举这个数据结构攻击ttt次

那么就相当于有ttt个空位（最后一个位置也是空位，但最后一个数据结构不是，需要单独考虑），插kkk个板，就是(tk)binom{t}{k}(kt)

发现除了k=0k=0k=0之外，都是是杨辉三角一列的之和，就等于(bi+1k+1)binom{b_i + 1}{k + 1}(k+1bi+1)

因为bib_ibi很大，不能直接算，但是kkk比较小，所以可以先O(1)O(1)O(1)计算出k=1k=1k=1时的值，然后O(1)O(1)O(1)递推下一个kkk的值

由于只有n−1n-1n−1个板，所以多项式长度始终不超过n−1n-1n−1。直接做多项式快速幂即可，再把mmm个多项式依次合起来

Code

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#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back
#define DEBUG(x) cout << #x << " = " << x << endl;

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int  (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 5e5 + 100;
const int Mod = 998244353;

namespace MATH
{
	inline void Add (int &a, int b) { if ((a += b) >= Mod) a -= Mod; }

	inline int Pow (int a, int b)
	{
		int ans = 1;
		for (int i = b; i; i >>= 1, a = (LL) a * a % Mod) if (i & 1) ans = (LL) ans * a % Mod;
		return ans;
	}
}

using namespace MATH;

int N, M;
int A[Maxn], B[Maxn];

namespace Poly
{
	int rev[Maxn], n;
	int _Wn[2][Maxn];

	inline void init ()
	{
		n = 5e5;
		for (int mid = 1; mid <= n; mid <<= 1)
		{
			_Wn[0][mid] = Pow (3, (Mod - 1) / (mid << 1));
			_Wn[1][mid] = Pow (_Wn[0][mid], Mod - 2);
		}
	}

	inline void dft (int *A, int fg)
	{
		for (int i = 0; i < n; ++i) if (rev[i] < i) swap (A[rev[i]], A[i]);
		for (int mid = 1; mid < n; mid <<= 1)
		{
			int Wn = _Wn[fg][mid], len = mid << 1;
			for (int i = 0; i < n; i += len)
				for (int j = i, W = 1; j < i + mid; ++j, W = (LL) W * Wn % Mod)
				{
					int x = A[j], y = (LL) W * A[j + mid] % Mod;
					A[j] = (x + y) % Mod;
					A[j + mid] = (x - y + Mod) % Mod;
				}
		}
		if (fg) for (int i = 0, inv = Pow (n, Mod - 2); i < n; ++i) A[i] = (LL) A[i] * inv % Mod;
	}

	inline void mul (int *A, int *B, int *C, int N)
	{
		for (n = 1; n <= (N << 1); n <<= 1);
		for (int i = 0; i < n; ++i) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) ? (n >> 1) : 0);

		static int F[Maxn], G[Maxn];
		for (int i = 0; i < n; ++i) F[i] = (i <= N) ? A[i] : 0;
		for (int i = 0; i < n; ++i) G[i] = (i <= N) ? B[i] : 0;

		dft (F, 0), dft (G, 0);
		for (int i = 0; i < n; ++i) F[i] = (LL) F[i] * G[i] % Mod;
		dft (F, 1);

		for (int i = 0; i <= (N << 1); ++i) C[i] = F[i];
	}
}

int F[Maxn], G[Maxn]; 
int H[Maxn];

inline void Solve ()
{
	++M;
	A[M] = 1, B[M] = B[M - 1] - 1;
	--A[M - 1];

	G[0] = 1;
	for (int i = 1; i <= M; ++i)
	{
		if (!A[i]) continue;
#define n (B[i] + 1)
#define m (j + 1)
		F[0] = (i == M) ? n : (n - 1);

		int res = (LL) n * (n - 1) / 2 % Mod;
		for (int j = 1; j <= N; ++j)
		{
			F[j] = res;
			res = (LL) res * (n - m) % Mod * Pow (m + 1, Mod - 2) % Mod;
		}

		for (int j = 0; j <= N; ++j) H[j] = 0;
		H[0] = 1;

		for (int j = A[i]; j; j >>= 1, Poly :: mul (F, F, F, N)) 
			if (j & 1) 
				Poly :: mul (H, F, H, N);

		Poly :: mul (G, H, G, N);
#undef n
#undef m
	}

	cout << G[N] << endl;
}

inline void Input ()
{
	N = read<int>() - 1, M = read<int>();
	for (int i = 1; i <= M; ++i) A[i] = read<int>(), B[i] = read<int>();
}

int main()
{

#ifdef hk_cnyali
	freopen("E.in", "r", stdin);
	freopen("E.out", "w", stdout);
#endif

	Poly :: init ();
	Input ();
	Solve ();

	return 0;
}

Farewell

Description

有一张 nnn个点 mmm条边的图，第 iii条边 ui,viu_i,v_iui,vi有 13frac{1}{3}31的概率从ui u_iui指向 viv_ivi ，另 13frac{1}{3}31 的概率从 viv_ivi 指向 uiu_iui ，剩下 13frac{1}{3}31的概率被删除。求这张图是有向无环图的概率

n≤20nle 20n≤20

Solution

[FWT] [子集卷积] [动态规划] [状态压缩]

设FSF_SFS表示SSS是DAG的方案数，ESE_SES表示点集SSS内部的边数，ES,TE_{S, T}ES,T表示SSS与TTT之间的边数

DAG计数显然枚举入度为000的点容斥

FS=∑T⊆S,T≠∅(−1)∣T∣+1FS−T×2ET,S−T
F_S = sum_{Tsubseteq S, Tne emptyset} (-1)^{|T| + 1}F_{S - T}times 2^{E_{T, S - T}}
FS=T⊆S,T≠∅∑(−1)∣T∣+1FS−T×2ET,S−T

2ET,S−T 2^{E_{T, S - T}}2ET,S−T是因为从S−TS-T S−T到T TT的边只能断掉或指向TTT

又因为2ET,S−T=2ES−ET−ES−T 2^{E_{T, S - T}} = 2^{E_S - E_T - E_{S - T}}2ET,S−T=2ES−ET−ES−T，所以式子可以化为

FS2ES=∑T⊆S,T≠∅(−1)∣T∣+12ET×FS−T2ES−T
frac{F_S}{2^{E_S}} = sum_{Tsubseteq S, Tne emptyset} frac{(-1)^{|T| + 1}}{2^{E_{T}}} times frac{F_{S - T}}{2^{E_{S - T}}}
2ESFS=T⊆S,T≠∅∑2ET(−1)∣T∣+1×2ES−TFS−T

子集卷积即可

Code


#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back
#define DEBUG(x) cout << #x << " = " << x << endl;

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int  (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 20 + 5, Maxs = (1 << 20) + 5;
const int Mod = 998244353;

namespace MATH
{
	inline void Add (int &a, int b) { if ((a += b) >= Mod) a -= Mod; }

	inline int Pow (int a, int b)
	{
		int ans = 1;
		for (int i = b; i; i >>= 1, a = (LL) a * a % Mod) if (i & 1) ans = (LL) ans * a % Mod;
		return ans;
	}
}

using namespace MATH;

int N, M, ALL;
int A[Maxs];
int E[Maxs];
int f[Maxn][Maxs], g[Maxn][Maxs];
int pw[Maxn * Maxn];

inline void Init ()
{
	ALL = (1 << N) - 1;
	pw[0] = 1;
	for (int i = 1; i <= M; ++i) pw[i] = (LL) pw[i - 1] * (Mod + 1) / 2 % Mod;

	for (int i = 1; i <= ALL; ++i)
	{
		int p = i & (-i);
		E[i] = E[i ^ p] + __builtin_popcount (A[p] & i);

		int len = __builtin_popcount (i);
		g[len][i] = (LL) ((len & 1) ? 1 : (Mod - 1)) * pw[E[i]] % Mod;
	}
}

inline void DWT (int *A, int n, int op)
{
	for (int mid = 1; mid < n; mid <<= 1)
		for (int i = 0, len = mid << 1; i < n; i += len)
			for (int j = i; j < i + mid; ++j)
			{
				int x = A[j], y = A[j + mid];
				if (!op) A[j + mid] = (x + y) % Mod;
				else A[j + mid] = (y - x + Mod) % Mod;
			}
}

inline void Solve ()
{
	Init ();

	f[0][0] = 1;
	DWT (f[0], 1 << N, 0);
	for (int i = 0; i <= N; ++i) DWT (g[i], 1 << N, 0);

	for (int i = 1; i <= N; ++i)
		for (int j = 0; j < i; ++j)
			for (int S = 0; S <= ALL; ++S)
				Add (f[i][S], (LL) f[j][S] * g[i - j][S] % Mod);

	DWT (f[N], 1 << N, 1);

	cout << (LL) f[N][ALL] * Pow (2, M) % Mod * Pow (Pow (3, M), Mod - 2) % Mod << endl; 
}

inline void Input ()
{
	N = read<int>(), M = read<int>();

	for (int i = 1; i <= M; ++i)
	{
		int x = read<int>() - 1, y = read<int>() - 1;
		A[1 << x] |= (1 << y);
		A[1 << y] |= (1 << x);
	}
}

int main()
{

#ifdef hk_cnyali
	freopen("F.in", "r", stdin);
	freopen("F.out", "w", stdout);
#endif

	Input ();
	Solve ();

	return 0;
}