D. Cycle in Graph
time limit per test
memory limit per test
input
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
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
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
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 1000000007
using 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;
}
}
}