# PAT(A) 1122. Hamiltonian Cycle (25)

### 1122. 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< 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 ... 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

### 解题报告

“Hamiltonian cycle”需要满足以下条件： 1. 从头到尾能走通，即存在路径。 2. 包含所有顶点。 3. 首尾一致。 4. 除首尾外，无多余重复顶点，即路径上定点数为图顶点数+1

### 代码

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 /* * Problem: 1122. Hamiltonian Cycle (25) * Author: HQ * Time: 2018-03-12 * State: Done * Memo: set */ #include "iostream" #include "set" using namespace std; int G[200 + 5][200 + 5]; int N, M,K; int main() { cin >> N >> M; int x, y; for (int i = 0; i < M; i++) { cin >> x >> y; G[x][y] = 1; G[y][x] = 1; } cin >> K; set route; for (int i = 0; i < K; i++) { int n; route.clear(); cin >> n >> x; int start = x; int flag = true; route.insert(x); for (int j = 1; j < n; j++) { cin >> y; if (!G[x][y]) flag = false; x = y; route.insert(y); } if (flag && route.size() == N && start == y && n == N + 1) cout << "YES" << endl; else cout << "NO" << endl; } system("pause"); }