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), 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:

6 106 23 41 52 53 14 11 66 31 24 567 5 1 4 3 6 2 56 5 1 4 3 6 29 6 2 1 6 3 4 5 2 64 1 2 5 17 6 1 3 4 5 2 67 6 1 2 5 4 3 1

Sample Output:

YESNONONOYESNO 第一，遍历点K一定等于N+1，因为既要遍历且一次所有顶点，而且回到起点 第二，遍历的最后一个点一定是起点 使用visit记录顶点是否遍历过了，graph记录路径是否能行

1 #include 
     
       2 #include 
      
        3 using namespace std; 4 int main() 5 { 6     int n, m, k, a, b, start, graph[205][205]; 7     bool visit[205];//记录每个顶点遍历一次 8     fill(graph[0], graph[0] + 25 * 205, -1); 9     cin >> n >> m;10     while (m--)11     {12         cin >> a >> b;13         graph[a][b] = graph[b][a] = 1;14     }15     cin >> m;16     while (m--)17     {18         cin >> k;19         bool flag = true;20         fill(visit, visit + 205, false);21         for (int i = 0; i < k; ++i)22         {23             cin >> b;24             if (flag == false || k != n + 1)//遍历所有的顶点并回到起点，则一定走过n+1个点25             {26                 flag = false;27                 continue;28             }29             if (i == 0)30                 start = b;//记录起点31             else if (graph[a][b] != 1)//此路不通32                 flag = false;33             else if (i == k - 1 && b != start)//最后一个点不是起点34                 flag = false;35             else if (i != k - 1 && visit[b] != false)//除了最后一次重复遍历起点，出现了其他点重复遍历，36                 flag = false;37             else38                 visit[b] = true;//遍历过39             a = b;//记录前一个点40         }41         if (flag)42         {43             for (int i = 1; i <= n && flag == true; ++i)44                 if (visit[i] == false)//存在没有遍历的顶点45                     flag = false;46             if (flag)47                 cout << "YES" << endl;48         }49         if (flag == false)50             cout << "NO" << endl;51     }52     return 0;    53 }