1122 Hamiltonian Cycle（25 分）

The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle".

In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2<N≤200), the number of vertices, and M, the number of edges in an undirected graph. Then M lines follow, each describes an edge in the format Vertex1 Vertex2, where the vertices are numbered from 1 to N. The next line gives a positive integer K which is the number of queries, followed by K lines of queries, each in the format:

n V​1​​ V​2​​ ... V​n​​

where n is the number of vertices in the list, and V​i​​'s are the vertices on a path.

Output Specification:

For each query, print in a line YES if the path does form a Hamiltonian cycle, or NO if not.

Sample Input:

6 10
6 2
3 4
1 5
2 5
3 1
4 1
1 6
6 3
1 2
4 5
6
7 5 1 4 3 6 2 5
6 5 1 4 3 6 2
9 6 2 1 6 3 4 5 2 6
4 1 2 5 1
7 6 1 3 4 5 2 6
7 6 1 2 5 4 3 1

Sample Output:

YES
NO
NO
NO
YES
NO

#include<cstdio>
#include<fstream>
#include<cstring>
#include<vector>
#include<set>
using namespace std;
const int maxn=210;

int n, m;
int g[maxn][maxn]={0};

int main(){
//	freopen("d://in.txt", "r", stdin);
	scanf("%d%d", &n, &m);
	for(int i=0; i<m; i++){
		int u, v;
		scanf("%d%d", &u, &v);
		g[u][v]=g[v][u]=1;
	}
	
	int k;
	scanf("%d", &k);
	for(int i=0; i<k; i++){
		int len;
		scanf("%d", &len);
		int ans[len+1]={0};
		set<int> s;
		for(int j=0; j<len; j++){
			scanf("%d", &ans[j]);
			s.insert(ans[j]); 
		}
		
		int x=ans[0];
		bool flag=true;
		if(x!=ans[len-1] || len!=n+1 || s.size()!=n) flag=false;
		for(int j=0; j<len-1; j++){
			int a=ans[j];
			int b=ans[j+1];
			if(g[a][b]!=1){
				flag=false;
				break;
			}
		}
		if(flag==true) printf("YES\n");
		else printf("NO\n");
		s.clear();
	}
	return 0;
}