A simple cycle is a closed simple path, with no other repeated vertices or edges other than the starting and ending vertices. The length of a cycle is the number of vertices on it. Given an undirected graph G(V, E), you are to detect whether it contains a simple
cycle of length K. To make the problem easier, we only consider cases with small K here.

There are multiple test cases.

The first line will contain a positive integer T (T ≤ 10) meaning the number of test cases.

For each test case, the first line contains three positive integers N, M and K ( N ≤ 50, M ≤ 500, 3 ≤ K ≤ 7). N is the number of vertices of the graph, M is the number of edges and K is the length of the cycle desired. Next follow M lines, each line contains
two integers A and B, describing an undirected edge AB of the graph. Vertices are numbered from 0 to N-1.

For each test case, you should output “YES” in one line if there is a cycle of length K in the given graph, otherwise output “NO”.

#include <iostream>

#include <stdio.h>

#include <string.h>

#include <string>

#include <stack>

#include <queue>

#include <map>

#include <set>

#include <vector>

#include <math.h>

#include <bitset>

#include <list>

#include <algorithm>

#include <climits>

using namespace std;

#define lson 2*i

#define rson 2*i+1

#define LS l,mid,lson

#define RS mid+1,r,rson

#define UP(i,x,y) for(i=x;i<=y;i++)

#define DOWN(i,x,y) for(i=x;i>=y;i--)

#define MEM(a,x) memset(a,x,sizeof(a))

#define W(a) while(a)

#define gcd(a,b) __gcd(a,b)

#define LL long long

#define N 200005

#define INF 0x3f3f3f3f

#define EXP 1e-8

#define lowbit(x) (x&-x)

const int mod = 1e9+7;

vector<int> a[550];

int vis[550],flag;

int n,m,k;

void dfs(int now,int pos,int pre)

{

    if(vis[now])

    {

        if(pos-vis[now]==k)

            flag = 1;

        return;

    }

    if(flag)

        return;

    vis[now]=pos;

    int i,len = a[now].size();

    for(i = 0; i<len; i++)

    {

        if(a[now][i]!=pre)

            dfs(a[now][i],pos+1,now);

    }

}

int main()

{

    int i,j,x,y,t;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d%d%d",&n,&m,&k);

        for(i = 0; i<=n; i++)

            a[i].clear();

        flag = 0;

        while(m--)

        {

            scanf("%d%d",&x,&y);

            a[x].push_back(y);

            a[y].push_back(x);

        }

        for(i=0; i<n; i++)

        {

            MEM(vis,0);

            dfs(i,1,-1);

        }

        printf("%s\n",flag?"YES":"NO");

    }

    return 0;

}