Codeforces 263 D. Cycle in Graph 环

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, ..., v_d such, that nodes v₁ and v_d are connected by an edge of the graph, also for any integer i (1 ≤ i < d) nodes v_i and v_i + 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 integersa_i, b_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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, ..., v_r (1 ≤ v_i ≤ 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 

找到一个长度大于k的环

#include <iostream>
#include <vector>
#include <cstring>
using namespace std;

vector<int>a[111111];
vector<int>path;
bool v[111111]={0};
bool done[111111]={0};
int dis[111111]={0};
int n,m,k;
bool ok=false;

void dfs(int i,int deep)
{
    if (v[i]) return;
    v[i]=true;
    path.push_back(i);
    dis[i]=deep;
    for (int j=0;j<a[i].size()&&!ok;j++)
    {
        if (v[a[i][j]]&&!done[a[i][j]]&&deep-dis[a[i][j]]>=k)
        {
            int pos=0;
            while (path[pos]!=a[i][j]) pos++;
            cout<<deep-pos+1<<endl;
            for (int k=pos;k<=deep;k++)
            {
                cout<<path[k]<<" ";
            }
            cout<<endl;
            ok=true;
        }
        else
        {
            dfs(a[i][j],deep+1);
        }
    }
    done[i]=true;
    path.pop_back();
}

int main()
{
    cin>>n>>m>>k;
    for (int i=1;i<=m;i++)
    {
        int x,y;
        cin>>x>>y;
        a[x].push_back(y);
        a[y].push_back(x);
    }
    dfs(1,0);
    return 0;
}