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 V₁ V₂ ... 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

思路

图中从一个点出发的一条路径能够走过所有的点并回到出发点，除起始点外所有其他点只能访问一次，这种情况产生的回路叫哈密尔顿回路。

所以验证输入的路径是不是哈密尔顿回路，必须满足以下条件：
1.输入的节点个数必须等于 总结点数 + 1
2.不能有重复出现的节点（只能走一次，起点除外）
3.起点终点必须相同。
4.两个节点之间必须直接相通（即被一条直线直接相连）

代码

#include<iostream>

#include<vector>

#include<set>

using namespace std;

vector<vector<int>> graph(201,vector<int>(201,-1));

int main()

{

    int N,M;

    while(cin >> N >> M )

    {

        for(int i = 1;i <= M;i++)

        {

            int a,b;

            cin >> a >> b;

            graph[a][b] = graph[b][a] = 1;

        }

        int K;

        cin >> K;

        for(int i = 0;i < K;i++)

        {

            int n;

            set<int> visits;

            cin >> n;

            vector<int> nodes(n + 1);

            for(int j = 1;j <= n;j++)

            {

               cin >> nodes[j];

               visits.insert(nodes[j]);

            }

            if(n != N + 1 || nodes[1] != nodes[n] || visits.size() != N)

            {

                cout << "NO" << endl;

                continue;

            }

            bool isha = true;

            for(int j = 2;j <= n;j++)

            {

                if(graph[nodes[j]][nodes[j - 1]] != 1)

                {

                    isha = false;

                    break;

                }

            }

            if(isha)

                cout << "YES" << endl;

            else

                cout << "NO" << endl;

        }

    }

}