CF1067E Random Forest Rank(树形dp，概率与期望，线性代数)

CF1067E Random Forest Rank

Solution

考虑树的邻接矩阵的秩的意义，不难发现相当于每个点找一个“代表”，对于一个点$x$，它的“代表”为一个与$x$相邻的结点，要求保证所有点的“代表”不重复，求能找到“代表”的点的最大点集。

稍加思考可以发现这个最大点集等同于最大匹配的两倍，而森林的秩等同于所有树的秩的和，还是最大匹配的两倍。

于是我们要求的东西相当于每个点能够匹配的概率的和。

令$f_x$表示$x$匹配到一个儿子结点的概率。
当$x$的所有儿子结点被匹配时，$x$无法匹配。因此$x$匹配的概率为$1-\prod \frac{1+f_v}{2}$，树形dp即可。

时间复杂度$O(n)$。

Code

#include <bits/stdc++.h>

using namespace std;

template<typename T> inline bool upmin(T &x, T y) { return y < x ? x = y, 1 : 0; }
template<typename T> inline bool upmax(T &x, T y) { return x < y ? x = y, 1 : 0; }

#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se second

typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int, int> PR;
typedef vector<int> VI; 

const lod eps = 1e-9;
const lod pi = acos(-1);
const int oo = 1 << 30;
const ll loo = 1ll << 60;
const int mods = 998244353;
const int inv2 = (mods + 1) >> 1;
const int MAXN = 600005;
const int INF = 0x3f3f3f3f; //1061109567
/*--------------------------------------------------------------------*/

namespace FastIO{
	constexpr int SIZE = (1 << 21) + 1;
	int num = 0, f;
	char ibuf[SIZE], obuf[SIZE], que[65], *iS, *iT, *oS = obuf, *oT = obuf + SIZE - 1, c;
	#define gc() (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)), (iS == iT ? EOF : *iS ++)) : *iS ++)
	inline void flush() {
		fwrite(obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	inline void putc(char c) {
		*oS ++ = c;
		if (oS == oT) flush();
	}
	inline void getc(char &c) {
		for (c = gc(); !isalpha(c) && c != EOF; c = gc());
	}
	inline void reads(char *st) {
		char c;
		int n = 0;
		getc(st[++ n]);
		for (c = gc(); isalpha(c) ; c = gc()) st[++ n] = c;
	}
	template<class I>
	inline void read(I &x) {
		for (f = 1, c = gc(); c < '0' || c > '9' ; c = gc()) if (c == '-') f = -1;
		for (x = 0; c >= '0' && c <= '9' ; c = gc()) x = (x << 3) + (x << 1) + (c & 15);
		x *= f;
	}
	template<class I>
	inline void print(I x) {
		if (x < 0) putc('-'), x = -x;
		if (!x) putc('0');
		while (x) que[++ num] = x % 10 + 48, x /= 10;
		while (num) putc(que[num --]);
	}
	struct Flusher_{~Flusher_(){flush();}} io_Flusher_;
}
using FastIO :: read;
using FastIO :: putc;
using FastIO :: reads;
using FastIO :: print;




vector<int> e[MAXN];
int f[MAXN], ans = 0;
int upd(int x, int y) { return x + y >= mods ? x + y - mods : x + y; }
void tree_dp(int x, int father) {
	for (auto v : e[x]) {
		if (v == father) continue;
		tree_dp(v, x);
	}
	int p = 1;
	for (auto v : e[x]) {
		if (v == father) continue;
		p = 1ll * p * (1 + f[v]) % mods * inv2 % mods;
	}
	f[x] = upd(1, mods - p);
	ans = upd(ans, f[x]);
}
signed main() {
#ifndef ONLINE_JUDGE
	freopen("a.in", "r", stdin);
#endif
	int n, pw = 1;
	read(n);
	for (int i = 1, u, v; i < n ; ++ i) read(u), read(v), e[u].PB(v), e[v].PB(u), pw = upd(pw, pw);
	tree_dp(1, 0);
	print(2ll * ans * pw % mods), putc('\n');
	return 0;
}