AtCoder Beginner Contest 223 G Vertex Deletion

洛谷传送门

AtCoder 传送门

设 \(f_{u,0/1}\) 为 \(u\) 的子树，\(u\) 是否在匹配内的最大匹配数。

注意到对于一个匹配，在它深度较浅的点上才会被计入答案。

转移大概是 \(f_{u,0}\) 取 \(\sum\limits_{v \in son_u} \max(f_{v,0}, f_{v,1})\)，\(f_{u,1}\) 要从儿子中选一个 \(f_{v,0}\)，剩下的选 \(\max(f_{v,0}, f_{v,1})\)。先假设全部选 \(\max(f_{v,0}, f_{v,1})\)，求出 \(\max\limits_{v \in son_u} f_{v,0} - \max(f_{v,0}, f_{v,1})\)，加进 \(f_{u,1}\) 即可。

发现还要求以父亲为根的子树的最大匹配，这个是换根基础操作，转移式同上。减去儿子的 \(\max\) 部分，维护前缀 \(\max\) 和后缀 \(\max\) 即可。注意特殊处理根。

code

// Problem: G - Vertex Deletion
// Contest: AtCoder - AtCoder Beginner Contest 223
// URL: https://atcoder.jp/contests/abc223/tasks/abc223_g
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 200100;
const int inf = 0x3f3f3f3f;

int n, head[maxn], len, ans, f[maxn][2], g[maxn][2];

struct edge {
	int to, next;
} edges[maxn << 1];

inline void add_edge(int u, int v) {
	edges[++len].to = v;
	edges[len].next = head[u];
	head[u] = len;
}

void dfs(int u, int fa) {
	int d = -inf;
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		if (v == fa) {
			continue;
		}
		dfs(v, u);
		f[u][0] += max(f[v][0], f[v][1]);
		f[u][1] += max(f[v][0], f[v][1]);
		d = max(d, f[v][0] - max(f[v][0], f[v][1]));
	}
	f[u][1] += d + 1;
}

void dfs2(int u, int fa, int f0, int f1) {
	int nf0 = f0, nf1 = f1;
	vector<int> son(1, 0);
	int k = max(f0, f1);
	f0 = f1 = max(f0, f1);
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		if (v == fa) {
			continue;
		}
		son.pb(v);
		k += max(f[v][0], f[v][1]);
		f0 += max(f[v][0], f[v][1]);
		f1 += max(f[v][0], f[v][1]);
	}
	if (k == max(f[1][0], f[1][1])) {
		++ans;
	}
	int cnt = (int)son.size() - 1;
	vector<int> pre(cnt + 2), suf(cnt + 2);
	pre[0] = suf[cnt + 1] = (u == 1 ? -inf : nf0 - max(nf0, nf1));
	for (int i = 1; i <= cnt; ++i) {
		int v = son[i];
		pre[i] = max(pre[i - 1], f[v][0] - max(f[v][0], f[v][1]));
	}
	for (int i = cnt; i; --i) {
		int v = son[i];
		suf[i] = max(suf[i + 1], f[v][0] - max(f[v][0], f[v][1]));
	}
	for (int i = 1; i <= cnt; ++i) {
		int v = son[i];
		if (u == 1 && cnt == 1) {
			dfs2(v, u, 0, -inf);
		} else {
			dfs2(v, u, f0 - max(f[v][0], f[v][1]), f1 - max(f[v][0], f[v][1]) + max(pre[i - 1], suf[i + 1]) + 1);
		}
	}
}

void solve() {
	scanf("%d", &n);
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v);
		add_edge(u, v);
		add_edge(v, u);
	}
	dfs(1, -1);
	dfs2(1, -1, 0, -inf);
	// printf("%d\n", max(f[1][0], f[1][1]));
	printf("%d\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}