大意: 無向圖, 無重邊自環, 每個點度數>=3, 要求完成下面任意一個任務

• 找一條結點數不少於n/k的簡單路徑
• 找k個簡單環, 每個環結點數小於n/k, 且不為3的倍數, 且每個環有一個特殊點$x$, $x$只屬於這一個環

任選一棵生成樹, 若高度>=n/k, 直接完成任務1, 否則對於葉子數一定不少於k, 而葉子反向邊數>=2, 一定可以構造出一個環

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head


const int N = 1e6+10;
int n, m, k;
vector<int> g[N];
int q[N];
int dep[N], vis[N], fa[N];

void dfs(int x, int f, int d) {
	vis[x]=1,fa[x]=f,dep[x]=d;
	if (d>=(n+k-1)/k) {
		puts("PATH");
		printf("%d\n", d);
		while (x) printf("%d ",x),x=fa[x];
		puts(""),exit(0);
	}
	int ok = 0;
	for (int y:g[x]) if (!vis[y]) {
		ok = 1, dfs(y,x,d+1);
	}
	if (!ok) q[++*q]=x;
}

int main() {
	scanf("%d%d%d", &n, &m, &k);
	REP(i,1,m) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].pb(v),g[v].pb(u);
	}
	dfs(1,0,1);
	puts("CYCLES");
	REP(i,1,k) {
		int x=0, y=0, z=q[i];
		for (int t:g[z]) if (t!=fa[z]) {
			if (!x) x=t;
			else y=t;
		}
		if ((dep[z]-dep[x]+1)%3) {
			printf("%d\n",dep[z]-dep[x]+1);
			for (; printf("%d ",z), z!=x; z=fa[z]) ;
		} else if ((dep[z]-dep[y]+1)%3) {
			printf("%d\n",dep[z]-dep[y]+1);
			for (; printf("%d ",z), z!=y; z=fa[z]) ;
		} else {
			if (dep[x]<dep[y]) swap(x,y);
			printf("%d\n", dep[x]-dep[y]+2);
			for (; printf("%d ",x), x!=y; x=fa[x]) ;
			printf("%d", z);
		}
		puts("");
	}
}

Johnny Solving CodeForces - 1103C (構造,圖論)