AGC029F

题意

给 \(n\ -\ 1\) 个集合 \(S_1,\ S_2,\ ...\ S_{n-1}\)，每个集合选两个数，连边，使得最后形成一棵树。

\(2\ \leq\ n\ \leq\ 10^5\)

做法1

考虑不合法，当且仅当存在一个集合 \(T\ =\ \{T_1,\ T_2,\ ...,\ T_k\}\) 使得 \(|S_{T_1}\ \cup\ S_{T_2}\ \cup\ S_{T_3}\ \cup\ ...\ \cup\ S_{T_k}|\ <\ k\)。否则合法。
用 Hall 定理判断，得到一组 \(n\ -\ 1\) 的匹配。令剩余的一个点为根，对其跑匈牙利算法，就能得到一棵树。由于交错路，所以可以证明合法。

代码

#include <bits/stdc++.h>

#ifdef __WIN32
#define LLFORMAT "I64"
#else
#define LLFORMAT "ll"
#endif

using namespace std;

struct edge {
	int to, cap, rev;

	edge() {}
	edge(int to, int cap, int rev): to(to), cap(cap), rev(rev) {}
};

struct max_flow {
	int n, S, T;
	vector<vector<edge> > g;

	max_flow() {}
	max_flow(int N): n(N) { g.resize(N); }
	max_flow(int N, int s, int t): n(N), S(s), T(t) { g.resize(N); }

	inline void add_edge(int from, int to, int cap) {
		g[from].push_back(edge(to, cap, g[to].size()));
		g[to].push_back(edge(from, 0, g[from].size() - 1));
		return;
	}

	inline int dinic() {
		vector<int> dst(n), iter(n);
		auto bfs = [&]() {
			fill(dst.begin(), dst.end(), -1);
			queue<int> q;
			q.push(S);
			dst[S] = 0;
			while(!q.empty()) {
				int u = q.front();
				q.pop();
				for (auto &e: g[u]) if(e.cap && !~dst[e.to]) dst[e.to] = dst[u] + 1, q.push(e.to);
			}
			return dst[T] != -1;
		};
		int ret = 0;
		while(bfs()) {
			fill(iter.begin(), iter.end(), 0);
			function<int(int, int)> dfs = [&](int u, int f) {
				if(u == T) return f;
				int t = f;
				for (int &i = iter[u]; i < g[u].size(); ++i) {
					auto &e = g[u][i];
					if(e.cap && dst[e.to] == dst[u] + 1) {
						int F = dfs(e.to, min(t, e.cap));
						e.cap -= F;
						g[e.to][e.rev].cap += F;
						t -= F;
						if(!t) return f;
					}
				}
				return f - t;
			};
			static const int oo = INT_MAX;
			for (int f = dfs(S, oo); f; f = dfs(S, oo)) ret += f;
		}
		return ret;
	}
};

int main() {
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<vector<int> > apr(n + 1);
	max_flow flow(n * 2, 0, 1);
	for (int i = 1; i < n; ++i) {
		int sz, foo = i + n;
		cin >> sz;
		flow.add_edge(foo, 1, 1);
		flow.add_edge(0, i + 1, 1);
		while(sz--) {
			int u;
			cin >> u;
			flow.add_edge(u, foo, 1);
			apr[u].push_back(foo);
		}
	}
	if(flow.dinic() != n - 1) { cout << "-1\n"; return 0; }
	vector<int> rem(n * 2, -1), con(n * 2, -1);
	for (int i = 1; i < n; ++i) {
		int u = i + n;
		for (auto &e: flow.g[u]) if(e.cap && e.to >= 2 && e.to <= n) rem[u] = e.to;
		if(!~rem[u]) { cout << "-1\n"; return 0; }
	}
	queue<int> q;
	q.push(1);
	while(!q.empty()) {
		int u = q.front();
		q.pop();
		for (int i: apr[u]) if(!~con[i]) {
			q.push(rem[i]);
			con[i] = u;
		}
	}
	for (int i = 1; i < n; ++i) {
		int u = i + n;
		if(!~rem[u] || !~con[u]) { cout << "-1\n"; return 0; }
	}
	for (int i = 1; i < n; ++i) {
		int u = i + n;
		cout << rem[u] << ' ' << con[u] << endl;
	}
	return 0;
}