Description

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.

Input

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.

Output

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”.

Sample Input

2
6 8 4
0 1
1 2
2 0
3 4
4 5
5 3
1 3
2 4
4 4 3
0 1
1 2
2 3
3 0

Sample Output

YES
NO

HINT

Source


题意：
问在一个图里面是否能找到一个长度为k的环

思路：
直接搜索看点是否反复訪问

#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;
}