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 1 #include <iostream>
2 #include <vector>
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 else
38 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 }
原文地址：https://www.cnblogs.com/zzw1024/p/11470328.html