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

题目大意：

给出一个图，判断给定的路径是不是哈密尔顿路径

分析：

1.设置falg1 判断节点是否多走、少走、或走成环
2.设置flag2 判断这条路能不能走通
3.当falg1、flag2都为1时是哈密尔顿路径，否则不是

原文链接：https://blog.csdn.net/liuchuo/article/details/53572051

题解

#include <bits/stdc++.h>

using namespace std;
int n,m;
int G[210][210];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int a,b;
    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>a>>b;
        G[a][b]=1;
        G[b][a]=1;
    }
    int k,num;
    cin>>k;
    while(k--){
        bool flag1=true,flag2=true;
        cin>>num;
        vector<int> v(num);
        set<int> s;
        for(int i=0;i<num;i++){
            cin>>v[i];
            s.insert(v[i]);
        }
        if(s.size()!=n||n!=num-1||v[0]!=v[num-1]) flag1=false;
        for(int i=0;i<num-1;i++)
            if(G[v[i]][v[i+1]]==0) flag2=false;
        printf("%s",flag1 && flag2 ? "YES\n" : "NO\n");
    }
    return 0;
}