「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

$dp[i][j]$表示到第$i$天，当前有$j$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 $(a,b,c)$，其中 $0lt alt b lt c le n$，表示在第 $a,b,c$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。

$nle 2times10^6$

Solution

[计数] [树状数组]

定义长度不超过 $frac{n-1}{2}$ ，且不含重复颜色的段为合法的段。记 $pre_x$ 为以 为右端点的合法段最远的左端点，$nxt_x$ 为以 $x$ 为左端点的合法段最远的右端点。

先枚举题目中的$a$，那么$bin(a, nxt_a + 1]$。在确定了$a, b$的位置后，合法的$c$位于$(b, nxt_b + 1]$和$[pre_a, n]$的交集中

注意，这里的$pre_a$必须是大于$a$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$a$，把合法的$b$对应的$(b, nxt_b + 1]$这段区间在树状数组中+1；查询就直接查$[pre_a, n]$的区间和；在离开$a$的时候把$(a, nxt_a + 1]$区间-1

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 = 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 $m$种数据结构(可把数据结构想像成游戏中的种族)，第 $i$种有 $a_i$个，每个可给敌人造成至少 $1$ 次，至多 $b_i$次伤害。有$n$名敌人，每人承担至少一次伤害。求总情况数模 $998244353$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。

$n times mle 10^5, a_ile 10^5, b_i < 998244353$

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$n-1$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$F_i(x)$表示一个第$i$种数据结构插板方案的生成函数，它的$k$次项系数表示插$k$个板的方案数。那么$displaystyle [x^k]F_i^{a_i}(x)$就是在第$i$种里插$k$个板的方案数了

---

考虑如何求$F_i(x)$

第$k$项系数其实就是

$binom{1}{k} + binom{2}{k} + cdots + binom{b_i}{k}$

枚举这个数据结构攻击$t$次

那么就相当于有$t$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$k$个板，就是$binom{t}{k}$

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

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

---

由于只有$n-1$个板，所以多项式长度始终不超过$n-1$。直接做多项式快速幂即可，再把$m$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 $n$个点 $m$条边的图，第 $i$条边 $u_i,v_i$有 $frac{1}{3}$的概率从$u_i$指向 $v_i$ ，另 $frac{1}{3}$ 的概率从 $v_i$ 指向 $u_i$ ，剩下 $frac{1}{3}$的概率被删除。求这张图是有向无环图的概率

$nle 20$

Solution

[FWT] [子集卷积] [动态规划] [状态压缩]

设$F_S$表示$S$是DAG的方案数，$E_S$表示点集$S$内部的边数，$E_{S, T}$表示$S$与$T$之间的边数

DAG计数显然枚举入度为$0$的点容斥

$F_S = sum_{Tsubseteq S, Tne emptyset} (-1)^{|T| + 1}F_{S - T}times 2^{E_{T, S - T}}$

$2^{E_{T, S - T}}$是因为从$S-T$到$T$的边只能断掉或指向$T$

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

$frac{F_S}{2^{E_S}} = sum_{Tsubseteq S, Tne emptyset} frac{(-1)^{|T| + 1}}{2^{E_{T}}} times frac{F_{S - T}}{2^{E_{S - T}}}$

子集卷积即可

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 = 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; }