【gdoi2018 day2】第二题 滑稽子图（subgraph）（性质DP+多项式）

题目大意 【gdoi2018 day2】第二题 滑稽子图（subgraph）

• 给你一颗树\(T\)，以及一个常数\(K\)，对于\(T\)的点集\(V\)的子集\(S\).

• 定义\(f(S)\)为点集\(S\)的导出子图的边数（一条原树中的边只有两个端点都出现在\(S\)中，才会出现在导出子图中）

数据范围

解题方案

\(Part_1\) 5%

• 随便做

\(Part_2\) 30%

• 考虑一下DP.

• 设\(f[i][j][0/1]\)表示第\(i\)个点，导出子图边数为\(j\)，第\(i\)个点是否选.

• 转移讨论一下即可，这里需要注意一下转移时用顺推还是逆推.

• 一般来说，对于背包类型的，我们顺推可以使每一个转移都是有效的.

• 枚举之前子树大小和，以及当前子树大小，乘积的和是\(O(n^2)\)级别的.

• 证明：

  • 两个点会产生贡献，当且仅当它们的\(LCA\)为当前的根时，每两个点的\(LCA\)唯一，故只有\(O(n^2)\)个点会产生贡献.

\(Part_3\) 100% DP

• 依然是DP，但我们要利用好\(k\le 10\)这个特性.

• 我们来观察一下顺推时的转移，其实可以表示为：

  • 令当前计算的是\(y\)对于父亲\(x\)的贡献，表达如下：
  \[f[x][i + j] = \sum_{i = 0}^{size_x - 1}\sum_{j=0}^{size_y-1} f[x][i] * f[y][j] * (i + j) ^ k
  \]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

$$f[x][i + j] = \sum_{i = 0}^{size_x - 1}\sum_{j = 0}^{size_y - 1} f[x][i]f[y][j] * \sum_{t = 0}^kitj^{k-t}\binom{k}{t}$$

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

$$f[x][i + j] = \sum_{i = 0}^{size_x - 1}\sum_{j = 0}^{size_y - 1} \sum_{t = 0}^k\binom{k}{t}f[x][i]f[y][j] i^t*j{k-t}$$

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

$$f[x][i + j] = \sum_{i = 0}^{size_x - 1}\sum_{j = 0}^{size_y - 1} \sum_{t = 0}^{k\binom{k}{t}(f[x][i]*i}t) (f[y][j]j^{k-t})$$

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

$$f[x][i + j] = \sum_{t = 0}^k\binom{k}{t}\sum_{i = 0}^{size_x - 1}\sum_{j = 0}^{size_y - 1} (f[x][i]i^t) (f[y][j]j^{k-t})$$

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

$$f[x][i + j] = \sum_{t = 0}^k\binom{k}{t}\sum_{i = 0}^{size_x - 1} (f[x][i]i^t)\sum_{j = 0}^{size_y - 1} (f[y][j]j^{k-t})$$

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

$$令g[x][t]表示\sum_{i = 0}^{{size_x-1}f[x][i]*i}t，则上式转化为$$

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

$$f[x][i + j] = \sum_{t = 0}^k\binom{k}{t}g[x][t]*g[y][k-t]$$

• 我们发现实质上\(f\)的转移只与\(g\)有关，所以我们直接枚举指数\(k\)大小，\(O(k)\)转移即可.

• 上述讨论的是当\(x,y\)两点不都选的时候，如果都选，转移应当如下：

\[f[x][i + j + 1] = \sum_{i = 0}^{size_x - 1}\sum_{j=0}^{size_y-1} f[x][i] * f[y][j] * (i + j + 1) ^ k
\]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

\[f[x][i + j + 1] = \sum_{i = 0}^{size_x - 1}\sum_{j=0}^{size_y-1} f[x][i] * f[y][j] * (i + j + 1) ^ k
\]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

\[f[x][i + j + 1] = \sum_{i = 0}^{size_x - 1}\sum_{j=0}^{size_y-1} f[x][i] * f[y][j] * \sum_{t = 0}^k(i+j)^t*1^{k-t}*\binom{k}{t}
\]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

\[f[x][i + j + 1] = \sum_{i = 0}^{size_x - 1}\sum_{j=0}^{size_y-1}\sum_{t = 0}^k f[x][i] * f[y][j] * (i+j)^t*\binom{k}{t}
\]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

\[f[x][i + j + 1] = \sum_{t = 0}^k \binom{k}{t} \sum_{i = 0}^{size_x - 1}\sum_{j=0}^{size_y-1} f[x][i] * f[y][j] * (i+j)^t
\]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

\[f[x][i + j + 1] = \sum_{t = 0}^k \binom{k}{t} \sum_{i = 0}^{size_x - 1}\sum_{j=0}^{size_y-1} f[x][i] * f[y][j] * \sum_{p = 0}^ti^p*j^{t-p}*\binom{t}{p}
\]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

\[f[x][i + j + 1] = \sum_{t = 0}^k \binom{k}{t} \sum_{i = 0}^{size_x - 1}\sum_{j=0}^{size_y-1} \sum_{p = 0}^t \binom{t}{p}f[x][i] * f[y][j] * i^p*j^{t-p}
\]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

\[f[x][i + j + 1] = \sum_{t = 0}^k \binom{k}{t} \sum_{p = 0}^t \binom{t}{p}\sum_{i = 0}^{size_x - 1}\sum_{j=0}^{size_y-1} f[x][i] * f[y][j] * i^p*j^{t-p}
\]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

\[f[x][i + j + 1] = \sum_{t = 0}^k \binom{k}{t} \sum_{p = 0}^t \binom{t}{p}\sum_{i = 0}^{size_x - 1} (f[x][i] * i^p) \sum_{j=0}^{size_y-1} (f[y][j] *j^{t-p})
\]

\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Downarrow\)

\[f[x][i + j + 1] = \sum_{t = 0}^k \binom{k}{t} \sum_{p = 0}^t \binom{t}{p}g[x][p]*g[y][t-p]
\]

• 可以发现，最后的转移，依然是只与\(g\)有关.

• 转移一个点是\(O(k^2)\)的，\(n\)个点，故时间复杂度为\(O(n * k^3/6)\)，实际上可以优化到\(O(n*k^2)\)

• 注意，我们如果有\(a_1<a_2<\cdots a_n 且 (1\le\ a_i \le m)\)，时间复杂度是\(O(\frac{n^m}{n!})\)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cctype>

#define ll long long
#define I register int
#define L register ll
#define F(i, a, b) for (L i = a; i <= b; i ++)
#define mec(a, b) memcpy(a, b, sizeof a)
#define mem(a, b) memset(a, b, sizeof a)
#define add(a, b) ((a) = (a + b) % mo)
#define N 100100
#define M 2 * N
#define Get getchar()
#define mo 998244353

using namespace std;

ll n, m, K, u, v, ans, sum, h, x, k;
ll F[N][11][2], G[N][11][2], jc[N], ny[N], d[N], bz[N];
ll nex[M], tov[M], las[N], last[N], tot;

void R(L &x) {
	char c = Get; x = 0; L t = 1;
	for (; !isdigit(c); c = Get) t = (c == '-' ? -1 : t);
	for (; isdigit(c); x = (x << 3) + (x << 1) + c - '0', c = Get); x = x * t;
}

void W(L x) {
	if (x < 0) { putchar('-'); W(-x); return; }
	if (x > 9) W(x / 10); putchar(x % 10 + '0');
}

void ins(L x, L y) { tov[++ tot] = y, nex[tot] = las[x], las[x] = tot; }

ll ksm(L x, L y) {
	L ans = 1;
	while (y) {
		if (y & 1) ans = (ans * x) % mo;
		x = (x * x) % mo, y >>= 1;
	}
	return ans;
}

ll C(L x, L y) { return jc[x] * ny[y] % mo * ny[x - y] % mo; }

void Dg_dp() {
	bz[1] = 1, d[h = 1] = 1;
	while (h) { x = d[h], k = las[x];
		if (k) {
			if (!bz[tov[k]]) bz[tov[k]] = 1, d[++ h] = tov[k];
			las[x] = nex[k];
		}
		if (!k) {
			G[x][0][0] = G[x][0][1] = 1;
			for (L k = last[x], y, s; k; k = nex[k])
				if ((y = tov[k]) && (!bz[y])) {
					mem(F[x], 0);
					F(j, 0, K)
						F(k, 0, j) {
							add(F[x][j][0], G[x][k][0] * (G[y][j - k][0] + G[y][j - k][1]) % mo * C(j, k));
							add(F[x][j][1], G[x][k][1] * (G[y][j - k][0]) % mo * C(j, k)); s = 0;
							F(p, 0, k) add(s, G[x][p][1] * G[y][k - p][1] % mo * C(k, p));
							add(F[x][j][1], C(j, k) * s);
						}
				F(j, 0, K) G[x][j][0] = F[x][j][0], G[x][j][1] = F[x][j][1];
			}
			bz[d[h --]] = 0;
		}

	}
}

int main() {
	freopen("subgraph.in", "r", stdin);
	freopen("subgraph.out", "w", stdout);

	R(n), R(m), R(K), jc[0] = ny[0] = 1;
	F(i, 1, m) R(u), R(v), ins(u, v), ins(v, u);
	F(i, 1, K) jc[i] = (jc[i - 1] * i) % mo, ny[i] = ksm(jc[i], mo - 2);
	if (K == 0) { printf("%d", ksm(2, n)); return 0;}
	mec(last, las), Dg_dp(), W((G[1][K][0] + G[1][K][1]) % mo);
}