DPV Subtree

题意

给定一棵 \(n\) 个节点的线段树。

任意黑白染色，求每个点被染成黑色且黑色点组成连通块的方案数。

Sol

考虑换根dp，钦定当前点作为根节点。

• \(f_i\) 表示当前子树内的方案数。
• \(g_i\) 表示子树外的方案数。

\(f\) 的转移显然是 \(f_u = \prod f_v + 1\)。

考虑 \(g\) 的转移。\(fa\) 子树外的贡献，以及 \(x\) 的兄弟儿子的贡献。

对于前者固然为 \(g_{fa}\)，后者考虑预处理前缀积后缀积。

需要注意的是，两者都是只有当 \(fa\) 被选时才会有贡献，所以 \(g_{fa}\) 不 \(+1\)。

\(g_u = 1 + g_{fa} \times \prod f_v + 1, v \in {brother}\)。

Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#define int long long
using namespace std;
#ifdef ONLINE_JUDGE

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;

#endif
int read() {
	int p = 0, flg = 1;
	char c = getchar();
	while (c < '0' || c > '9') {
		if (c == '-') flg = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9') {
		p = p * 10 + c - '0';
		c = getchar();
	}
	return p * flg;
}
void write(int x) {
	if (x < 0) {
		x = -x;
		putchar('-');
	}
	if (x > 9) {
		write(x / 10);
	}
	putchar(x % 10 + '0');
}
const int N = 1e5 + 5, M = 2e5 + 5;
int mod;

namespace G {

array <int, N> fir;
array <int, M> nex, to;
int cnt;
void add(int x, int y) {
	cnt++;
	nex[cnt] = fir[x];
	to[cnt] = y;
	fir[x] = cnt;
}

}

array <int, N> f, g;

array <int, N> fa;
void dfs1(int x) {
	f[x] = 1;
	for (int i = G::fir[x]; i; i = G::nex[i]) {
		if (G::to[i] == fa[x]) continue;
		fa[G::to[i]] = x;
		dfs1(G::to[i]);
		f[x] = f[x] * (f[G::to[i]] + 1) % mod;
	}
}

array <int, N> tp;
void dfs2(int x) {
	int cnt = 0;
	for (int i = G::fir[x]; i; i = G::nex[i]) {
		if (G::to[i] == fa[x]) continue;
		cnt++, tp[cnt] = f[G::to[i]] + 1;
	}
	tp[cnt + 1] = 1;
	for (int i = cnt; i; i--)
		tp[i] = tp[i] * tp[i + 1] % mod;
	cnt = 0;
	int kp = 1;
	for (int i = G::fir[x]; i; i = G::nex[i]) {
		if (G::to[i] == fa[x]) continue;
		cnt++;
		g[G::to[i]] = g[x] * kp % mod * tp[cnt + 1] % mod + 1;
		kp = kp * (f[G::to[i]] + 1) % mod;
	}
	for (int i = G::fir[x]; i; i = G::nex[i]) {
		if (G::to[i] == fa[x]) continue;
		dfs2(G::to[i]);
	}

}

signed main() {
	int n = read();
	mod = read();
	for (int i = 2; i <= n; i++) {
		int x = read(), y = read();
		G::add(x, y), G::add(y, x);
	}
	g[1] = 1;
	dfs1(1), dfs2(1);
	for (int i = 1; i <= n; i++) write(f[i] * g[i] % mod), puts("");
	return 0;
}