Description：
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 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
n个点，m调路径，求长度不超过k+1的环，哈密尔顿回路判环，确定长度输出
AC代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<queue>
#include<set>
#include<iomanip>
#include<math.h>
using namespace std;
typedef long long ll;
typedef double ld;
const int INF = 0x3f3f3f3f;
vector<int>mp[150000];
int pos[150000],ans[150000];
int n,m,k,cnt,ss,ee,ok;
void dfs(int u)
{
if(ok==1)
return ;
pos[u]=++cnt;
ans[cnt]=u;
for(int i=0; i<mp[u].size(); i++)
{
int v=mp[u][i];
if(pos[v])
{
if(pos[u]-pos[v]+1>=k+1)
{
ok=1;
ss=v;
ee=u;
return ;
}
}
else dfs(v);
}
}
int main()
{
while(~scanf("%d %d %d",&n,&m,&k))
{
ok=0;
cnt=0;
memset(pos,0,sizeof(pos));
for(int i=1; i<=m; i++)
{
int x,y;
scanf("%d%d",&x,&y);
mp[x].push_back(y);
mp[y].push_back(x);
}
dfs(1);
printf("%d\n",pos[ee]-pos[ss]+1);
for(int i=pos[ss]; i<=pos[ee]; i++)
{
printf("%d ",ans[i]);
}
printf("\n");
}
}
