PAT 1122 Hamiltonian Cycle

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;
}