AtCoder Beginner Contest 302 Ex Ball Collector

洛谷传送门

AtCoder 传送门

考虑如果只询问一次怎么做。连边 \((a_i, b_i)\)，对于每个连通块分别考虑。这是 ARC111B，如果一个连通块是树，肯定有一个点不能被选；否则有环，一定能构造一种方案，使得每个点都被选。

那么现在对于每个点都要求，考虑 dfs 时合并当前的 \((a_u, b_u)\)，并且使用可撤销并查集。具体而言，把每次的修改都压进栈里，退出一个点就把这些修改全部复原。注意不要路径压缩，使用按秩合并。

code

// Problem: Ex - Ball Collector
// Contest: AtCoder - TOYOTA MOTOR CORPORATION Programming Contest 2023#2 (AtCoder Beginner Contest 302)
// URL: https://atcoder.jp/contests/abc302/tasks/abc302_h
// 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 long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 200100;

int n, a[maxn], b[maxn], fa[maxn], rnk[maxn], sz[maxn], e[maxn], top, ans, c[maxn];
pair<int*, int> stk[maxn * 50];
vector<int> G[maxn];

int find(int x) {
	return fa[x] == x ? x : find(fa[x]);
}

inline void merge(int x, int y) {
	x = find(x);
	y = find(y);
	stk[++top] = make_pair(&ans, ans);
	if (x == y) {
		ans -= (e[x] == sz[x] - 1 ? sz[x] - 1 : sz[x]);
		stk[++top] = make_pair(e + x, e[x]);
		++e[x];
		ans += (e[x] == sz[x] - 1 ? sz[x] - 1 : sz[x]);
		return;
	}
	ans -= (e[x] == sz[x] - 1 ? sz[x] - 1 : sz[x]);
	ans -= (e[y] == sz[y] - 1 ? sz[y] - 1 : sz[y]);
	if (rnk[x] <= rnk[y]) {
		stk[++top] = make_pair(fa + x, fa[x]);
		fa[x] = y;
		stk[++top] = make_pair(sz + y, sz[y]);
		sz[y] += sz[x];
		stk[++top] = make_pair(e + y, e[y]);
		e[y] += e[x] + 1;
		ans += (e[y] == sz[y] - 1 ? sz[y] - 1 : sz[y]);
		if (rnk[x] == rnk[y]) {
			stk[++top] = make_pair(rnk + y, rnk[y]);
			++rnk[y];
		}
	} else {
		stk[++top] = make_pair(fa + y, fa[y]);
		fa[y] = x;
		stk[++top] = make_pair(sz + x, sz[x]);
		sz[x] += sz[y];
		stk[++top] = make_pair(e + x, e[x]);
		e[x] += e[y] + 1;
		ans += (e[x] == sz[x] - 1 ? sz[x] - 1 : sz[x]);
	}
}

inline void undo() {
	*stk[top].fst = stk[top].scd;
	--top;
}

void dfs(int u, int fa) {
	int lsttop = top;
	merge(a[u], b[u]);
	c[u] = ans;
	for (int v : G[u]) {
		if (v == fa) {
			continue;
		}
		dfs(v, u);
	}
	while (top > lsttop) {
		undo();
	}
}

void solve() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d%d", &a[i], &b[i]);
		fa[i] = i;
		rnk[i] = sz[i] = 1;
	}
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v);
		G[u].pb(v);
		G[v].pb(u);
	}
	dfs(1, -1);
	for (int i = 2; i <= n; ++i) {
		printf("%d ", c[i]);
	}
}

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