CF1103C Johnny Solving 题解

考虑对此无向图建出 DFS 树，也就是 DFS 一次。

如果这棵树直径 $\geq \frac{n}{k}$，直接把路径找出来即可。

否则，我们找出所有叶子节点，树上必然至少有 $k$ 个叶子。由于每个。点度数大于等于 $3$，所以每个叶子必然有不小于两条返祖边。假设叶子节点为 $u$，其返祖边中任意两个点 $v,z$。由于是返祖边，显然 $u \rightarrow v$，$u \rightarrow z$，$v \rightarrow u \rightarrow z$ 都是环。我们设这三个点在 DFS 树上的深度分别为 $a,b,c$。那么，$u \rightarrow v$ 的环长度为 $a-b+1$，$u \rightarrow z$ 的环长度为 $a-c+1$，$v \rightarrow u \rightarrow z$ 的长度为 $b-c+2$。容易发现其中必然有一个不是 $3$ 的倍数。于是第一个限制满足了。

接着每个叶子节点作为那个只出现一次的点即可，就做完了。

#pragma GCC optimize("-Ofast,fast-math,-inline")
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <set>
#include <cassert>
#include <cstring>
#include <string>
#include <vector>
using namespace std;

const int N = 5e5 + 5;

int n, m, k;
vector<int> G[N];

int fa[N], dis[N];
bool f = 1;
int maxlen = -1, nowu = 0;
bool vv[N];

void dfs(int u, int f)
{
	vv[u] = 1;
	fa[u] = f;
	dis[u] = dis[f] + 1;
	if (dis[u] > maxlen)
	{
		maxlen = dis[u];
		nowu = u;
	}
	for (auto& j : G[u])
	{
		if (j != f && !vv[j])
		{
			dfs(j, u);
		}
	}
}

bool vis[N];
int cnt = 0;
int dep[N];
set<pair<int, int> > st;
int p = 0;

void ndfs(int u, int f)
{
	fa[u] = f;
	dep[u] = dep[f] + 1;
	vis[u] = 1;
	int c = 0;
	for (auto& j : G[u])
	{
		if (j == f) continue;
		if (!vis[j])
		{
			c++;
		}
	}
	int cc = 0;
	int X1 = 0, X2 = 0;
	for (auto& j : G[u])
	{
		if (j == f) continue;
		if (vis[j] && !c && !cc)
		{
			if (!X1) X1 = j;
			else X2 = j;
			if (st.count(make_pair(u, j))) continue;
			st.insert(make_pair(u, j));
			st.insert(make_pair(j, u));
			int len = dep[u] - dep[j] + 1;
			if (len % 3 == 0 || len < 3) continue;
			cnt++;
			cout << len << "\n";
			int ru = u;
			cc++;
			while (ru != j)
			{
				cout << ru << " ";
				ru = fa[ru];
			}
			cout << j << " ";
			cout << "\n";
			if (cnt == k)
			{
				exit(0);
			}
		}
		else if (!vis[j])
		{
			ndfs(j, u);
		}
	}
	if (!cc && !c)
	{
		if (dep[X1] < dep[X2]) swap(X1, X2);
		int len = dep[X1] - dep[X2] + 2;
		if (len % 3 != 0 && len > 3)
		{
			cout << len << "\n" << u << " ";
			int nowu = X1;
			while (nowu != X2)
			{
				cout << nowu << " ";
				nowu = fa[nowu];
			}
			cout << X2 << "\n";
			cnt++;
			if (cnt == k) exit(0);
		}
	}
}

int main()
{
	ios::sync_with_stdio(0), cin.tie(nullptr), cout.tie(nullptr);
	cin >> n >> m >> k;
	for (int i = 1; i <= m; i++)
	{
		int u, v;
		cin >> u >> v;
		G[u].emplace_back(v);
		G[v].emplace_back(u);
	}
	dfs(1, 0);
	memset(vv, 0, sizeof vv);
	f = 0;
	int p = nowu;
	maxlen = -1;
	nowu = 0;
	dfs(p, 0);
	if (maxlen >= (n - 1) / k + 1)
	{
		cout << "PATH\n";
		cout << maxlen << "\n";
		while (nowu != p)
		{
			cout << nowu << " ";
			nowu = fa[nowu];
		}
		cout << p << " ";
		cout << "\n";
		return 0;
	}
	memset(dep, 0, sizeof dep);
	memset(fa, 0, sizeof fa);
	memset(vis, 0, sizeof vis);
	st.clear();
	cout << "CYCLES\n";
	ndfs(1, 0);
	return 0;
}