codeforces 263D

D. Cycle in Graph

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You've got a undirected graph G, consisting of n nodes. We will consider the nodes of the graph indexed by integers from 1 to n. We know that each node of graph G is connected by edges with at least k other nodes of this graph. Your task is to find in the given graph a simple cycle of length of at least k + 1.

A simple cycle of length d (d > 1) in graph G is a sequence of distinct graph nodes v1, v2, ..., vd such, that nodes v1 and vd are connected by an edge of the graph, also for any integer i (1 ≤ i < d) nodes vi and vi + 1 are connected by an edge of the graph.

Input

The first line contains three integers n, m, k (3 ≤ n, m ≤ 105; 2 ≤ k ≤ n - 1) — the number of the nodes of the graph, the number of the graph's edges and the lower limit on the degree of the graph node. Next m lines contain pairs of integers. The i-th line contains integersai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — the indexes of the graph nodes that are connected by the i-th edge.

It is guaranteed that the given graph doesn't contain any multiple edges or self-loops. It is guaranteed that each node of the graph is connected by the edges with at least k other nodes of the graph.

Output

In the first line print integer r (r ≥ k + 1) — the length of the found cycle. In the next line print r distinct integers v1, v2, ..., vr (1 ≤ vi ≤ n)— the found simple cycle.

It is guaranteed that the answer exists. If there are multiple correct answers, you are allowed to print any of them.

Sample test(s)

input

3 3 2
1 2
2 3
3 1

output

3
1 2 3 

input

4 6 3
4 3
1 2
1 3
1 4
2 3
2 4

output

4
3 4 1 2 

环中点大于k....

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<string>
#include<vector>
#include<stack>
#include<set>
#include<queue>
#include<map>
using namespace std;
vector<int> e[100010];
int ans[100010],n,m,k,top,st,ed,Dfs[100010];
bool dfs(int u)
{
	Dfs[u]=++top;
	ans[top]=u;
	for(int i=0;i<e[u].size();i++)
	{
		int v=e[u][i];
		if(Dfs[v])
		{
			if(top-Dfs[v]<k) continue;
			st=Dfs[v];
			ed=Dfs[u];
			return true;
		}
		if(dfs(v))
			return true;
	}
	return false;
}

int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for(int i=0;i<m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs(1);
	printf("%d
",ed-st+1);
	for(int i=st;i<=ed;i++)
		printf("%d ",ans[i]);
	printf("
");
	return 0;
}