Codeforces Round #161 (Div. 2)-D. Cycle in Graph

原题链接

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 integers ai, 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.

Examples

input

3 3 21 22 33 1

output

31 2 3 

input

4 6 34 31 21 31 42 32 4

output

43 4 1 2 

从一个点开始深搜，记录每个节点的深度，若i和j有点相连，且d[i] - d[j] >= k则输出j到i的路径

#include <bits/stdc++.h>#define maxn 100005#define MOD 1000000007using namespace std;typedef long long ll;vector<int> v[maxn];int d[maxn], pre[maxn];int n, m, k;bool dfs(int j, int f){for(int i = 0; i < v[j].size(); i++){int h = v[j][i];if(h == f)continue;if(d[h] == 0){d[h] = d[j] + 1;pre[h] = j;if(dfs(h, j))return true;}else{if(d[h] < d[j] && d[j] - d[h] >= k){int s = d[j] - d[h] + 1;printf("%d\n", s);printf("%d", j);s--;while(s--){printf(" %d", pre[j]);j = pre[j];}return true;}}}return false;}int main(){//freopen("in.txt", "r", stdin);int a, b;scanf("%d%d%d", &n, &m, &k);for(int i = 0; i < m; i++){scanf("%d%d", &a, &b);v[a].push_back(b);v[b].push_back(a);}for(int i = 1; i <= n; i++){if(d[i] == 0){d[i] = 1;if(dfs(i, -1)) return 0;}}}