AtCoder Beginner Contest 327 D

AtCoder Beginner Contest 327D

D - Good Tuple Problem (atcoder.jp)(种类并查集,二分图染色)

算法学习笔记(7)：种类并查集

附上典题:P1892 [BOI2003] 团伙

种类并查集做法

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
template<typename T>
struct UFS {
	T sz;
	vector<T> rank, p;
	void link(T x, T y) {
		if (x == y)
			return;
		if (rank[x] > rank[y])
			p[y] = x;
		else
			p[x] = y;
		if (rank[x] == rank[y])
			rank[y]++;
	}
	void init(int n) {
		sz = n;
		rank.resize(n + 1);
		p.resize(n + 1);
		for (int i = 0; i <= sz; i++) {
			p[i] = i;
			rank[i] = 0;
		}
	}
	T find(T x) {
		return x == p[x] ? x : (p[x] = find(p[x]));
	}
	void unin(T x, T y) {
		link(find(x), find(y));
	}
	void compress() {
		for (int i = 0; i < sz; i++)
			find(i);
	}
};
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n, M;
	cin >> n >> M;
	vector<int> A(M), B(M);
	for (auto &i : A)
		cin >> i;
	for (auto &i : B)
		cin >> i;
 
	UFS<i64> ufs;
	ufs.init(n * 2);
	for (int i = 0; i < M; i ++) {
		int u = ufs.find(A[i]);
		int v = ufs.find(B[i]);
		if (u == v) {
			cout << "No\n";
			return 0;
		}
		ufs.unin(B[i] + n, u);
		ufs.unin(A[i] + n, v);
	}
 
	cout << "Yes\n";
 
	return 0;
}

二分图染色

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n, m;
	cin >> n >> m;
	vector G(n + 1, vector<int>());
	vector<int> co(n + 1, -1), A(m), B(m);
	for (auto &i : A) cin >> i;
	for (auto &i : B) cin >> i;
 
	for (int i = 0; i < m; i ++) {
		G[A[i]].push_back(B[i]);
		G[B[i]].push_back(A[i]);
	}
 
	auto dfs = [&](auto self, int x, int val) -> void{
		co[x] = val;
		for (auto i : G[x]) {
			if (co[i] == -1) {
				self(self, i, val ^ 1);
			} else if (co[i] == val) {
				cout << "No\n";
				exit(0);
			}
		}
	};
 
	for (int i = 1; i <= n; i ++)
		if (co[i] == -1)
			dfs(dfs, i, 0);
 
	cout << "Yes\n";
 
	return 0;
}