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 in 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 shocould output "YES" in one line if there is a cycle of length K in the given graph, otherwise output "NO ".
Sample Input
26 8 40 11 22 03 44 55 31 32 44 4 30 11 22 33 0
Sample Output
YESNO
HINT
Source
Q: Can I find a k-length ring in a graph?
Idea: directly search for repeated access points
#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