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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> CSU1660: K-Cycle

CSU1660: K-Cycle

編輯:C++入門知識

CSU1660: K-Cycle


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


題意: 問在一個圖裡面能否找到一個長度為k的環
思路: 直接搜索看點是否重復訪問
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
#include 
#include 
#include 
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 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


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