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 V2​ ... Vn

​​

where n is the number of vertices in the list, and Vi'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<iostream> //水题
#include<vector>
using namespace std;
int main(){
int vn, en, qn;
cin>>vn>>en;
vector<vector<int>> map(vn+1, vector<int>(vn+1, 0));
vector<int> visited(vn+1, 0);
for(int i=0; i<en; i++){
int v1, v2;
cin>>v1>>v2;
map[v1][v2]=map[v2][v1]=1;
}
cin>>qn;
for(int i=0; i<qn; i++){
int n, flag=0;
cin>>n;
vector<int> path(n, 0);
vector<vector<int>> temp=map;
vector<int> visited(vn+1, 0);
for(int j=0; j<n; j++)
cin>>path[j];
if(path[0]!=path[n-1]){
cout<<"NO"<<endl;
continue;
}
for(int j=0; j<n-1; j++)
if(temp[path[j]][path[j+1]]==1){
visited[path[j]]=visited[path[j+1]]=1;
temp[path[j]][path[j+1]]=temp[path[j+1]][path[j]]=0;
}else{
flag=1;
break;
}
for(int j=1; j<=vn; j++)
if(visited[j]!=1)
flag=1;
flag==0?cout<<"YES"<<endl:cout<<"NO"<<endl;
}
return 0;
}

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