PAT甲级——A1122 Hamiltonian Cycle【25】

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 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

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

 #include <iostream>
 #include <vector>
 using namespace std;
 int main()
 {
     int n, m, k, a, b, start, graph[][];
     bool visit[];//记录每个顶点遍历一次
     fill(graph[], graph[] +  * , -);
     cin >> n >> m;
     while (m--)
     {
         cin >> a >> b;
         graph[a][b] = graph[b][a] = ;
     }
     cin >> m;
     while (m--)
     {
         cin >> k;
         bool flag = true;
         fill(visit, visit + , false);
         for (int i = ; i < k; ++i)
         {
             cin >> b;
             if (flag == false || k != n + )//遍历所有的顶点并回到起点，则一定走过n+1个点
             {
                 flag = false;
                 continue;
             }
             if (i == )
                 start = b;//记录起点
             else if (graph[a][b] != )//此路不通
                 flag = false;
             else if (i == k -  && b != start)//最后一个点不是起点
                 flag = false;
             else if (i != k -  && 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，
                 flag = false;
             else
                 visit[b] = true;//遍历过
             a = b;//记录前一个点
         }
         if (flag)
         {
             for (int i = ; i <= n && flag == true; ++i)
                 if (visit[i] == false)//存在没有遍历的顶点
                     flag = false;
             if (flag)
                 cout << "YES" << endl;
         }
         if (flag == false)
             cout << "NO" << endl;
     }
     return ;
 }